261_A. Maxim and Discounts
Maxim always goes to the supermarket on Sundays. Today the supermarket has a special offer of discount systems.

There are m types of discounts. We assume that the discounts are indexed from 1 to m. To use the discount number i, the customer takes a special basket, where he puts exactly qi items he buys. Under the terms of the discount system, in addition to the items in the cart the customer can receive at most two items from the supermarket for free. The number of the "free items" (0, 1 or 2) to give is selected by the customer. The only condition imposed on the selected "free items" is as follows: each of them mustn't be more expensive than the cheapest item out of the qi items in the cart.

Maxim now needs to buy n items in the shop. Count the minimum sum of money that Maxim needs to buy them, if he use the discount system optimally well.

Please assume that the supermarket has enough carts for any actions. Maxim can use the same discount multiple times. Of course, Maxim can buy items without any discounts.

Input

The first line contains integer m (1 ≤ m ≤ 105) — the number of discount types. The second line contains m integers: q1, q2, ..., qm (1 ≤ qi ≤ 105). 

The third line contains integer n (1 ≤ n ≤ 105) — the number of items Maxim needs. The fourth line contains n integers: a1, a2, ..., an (1 ≤ ai ≤ 104) — the items' prices.

The numbers in the lines are separated by single spaces.

Output

In a single line print a single integer — the answer to the problem.

Examples

Input

1
2
4
50 50 100 100


Output

200


Input

2
2 3
5
50 50 50 50 50


Output

150


Input

1
1
7
1 1 1 1 1 1 1


Output

3

Note

In the first sample Maxim needs to buy two items that cost 100 and get a discount for two free items that cost 50. In that case, Maxim is going to pay 200.

In the second sample the best strategy for Maxim is to buy 3 items and get 2 items for free using the discount. In that case, Maxim is going to pay 150.
```python
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from decimal import *
from io import BytesIO, IOBase
from itertools import *
from collections import *

# sys.setrecursionlimit(10**5)
M = 10 ** 9 + 7

# print(math.factorial(5))
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)

# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

input = lambda: sys.stdin.readline().rstrip("\r\n")


def inp(): return sys.stdin.readline().rstrip("\r\n")  # for fast input


def out(var): sys.stdout.write(str(var))  # for fast output, always take string


def lis(): return list(map(int, inp().split()))


def stringlis(): return list(map(str, inp().split()))


def sep(): return map(int, inp().split())


def strsep(): return map(str, inp().split())


def fsep(): return map(float, inp().split())


def inpu(): return int(inp())


# -----------------------------------------------------------------

def regularbracket(t):
    p = 0
    for i in t:
        if i == "(":
            p += 1
        else:
            p -= 1
        if p < 0:
            return False
    else:
        if p > 0:
            return False
        else:
            return True


# -------------------------------------------------
def binarySearchCount(arr, n, key):
    left = 0
    right = n - 1

    count = 0

    while (left <= right):
        mid = int((right + left) / 2)

        # Check if middle element is
        # less than or equal to key
        if (arr[mid] <= key):
            count = mid + 1
            left = mid + 1
        # If key is smaller, ignore right half
        else:
            right = mid - 1
    return count


# ------------------------------reverse string(pallindrome)
def reverse1(string):
    pp = ""
    for i in string[::-1]:
        pp += i
    if pp == string:
        return True
    return False


# --------------------------------reverse list(paindrome)
def reverse2(list1):
    l = []
    for i in list1[::-1]:
        l.append(i)
    if l == list1:
        return True
    return False


def mex(list1):
    # list1 = sorted(list1)
    p = max(list1) + 1
    for i in range(len(list1)):
        if list1[i] != i:
            p = i
            break
    return p


def sumofdigits(n):
    n = str(n)
    s1 = 0
    for i in n:
        s1 += int(i)
    return s1


def perfect_square(n):
    s = math.sqrt(n)
    if s == int(s):
        return True
    return False


# -----------------------------roman
def roman_number(x):
    if x > 15999:
        return
    value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
    symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
    roman = ""
    i = 0
    while x > 0:
        div = x // value[i]
        x = x % value[i]
        while div:
            roman += symbol[i]
            div -= 1
        i += 1
    return roman


def soretd(s):
    for i in range(1, len(s)):
        if s[i - 1] > s[i]:
            return False
    return True


# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
    ct = 0
    for i in range(2, h + 1, 2):
        for j in range(2, w + 1, 2):
            ct += (h - i + 1) * (w - j + 1)
    return ct


def countrhombi2(h, w):
    return ((h * h) // 4) * ((w * w) // 4)


# ---------------------------------
def binpow(a, b):
    if b == 0:
        return 1
    else:
        res = binpow(a, b // 2)
    if b % 2 != 0:
        return res * res * a
    else:
        return res * res


# -------------------------------------------------------
def binpowmodulus(a, b, m):
    a %= m
    res = 1
    while (b > 0):
        if (b & 1):
            res = res * a % m
        a = a * a % m
        b >>= 1
    return res


# -------------------------------------------------------------
def coprime_to_n(n):
    result = n
    i = 2
    while (i * i <= n):
        if (n % i == 0):
            while (n % i == 0):
                n //= i
            result -= result // i
        i += 1
    if (n > 1):
        result -= result // n
    return result


# -------------------prime
def prime(x):
    if x == 1:
        return False
    else:
        for i in range(2, int(math.sqrt(x)) + 1):
            # print(x)
            if (x % i == 0):
                return False
        else:
            return True

def luckynumwithequalnumberoffourandseven(x,n,a):
    if x >= n and str(x).count("4") == str(x).count("7"):
        a.append(x)
    else:
        if x < 1e12:
            luckynumwithequalnumberoffourandseven(x * 10 + 4,n,a)
            luckynumwithequalnumberoffourandseven(x * 10 + 7,n,a)
    return a
#----------------------
def luckynum(x,l,r,a):
    if x>=l and x<=r:
        a.append(x)
    if x>r:
        a.append(x)
        return a
    if x < 1e10:
            luckynum(x * 10 + 4, l,r,a)
            luckynum(x * 10 + 7, l,r,a)
    return a

def luckynuber(x, n, a):
    p = set(str(x))
    if len(p) <= 2:
        a.append(x)
    if x < n:
        luckynuber(x + 1, n, a)
    return a


# ------------------------------------------------------interactive problems

def interact(type, x):
    if type == "r":
        inp = input()
        return inp.strip()
    else:
        print(x, flush=True)


# ------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
    # Initialize result
    count = 0

    # Keep dividing n by
    # 5 & update Count
    while (n >= 5):
        n //= 5
        count += n

    return count


# -----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
    if len(arr) > 1:

        # Finding the mid of the array
        mid = len(arr) // 2

        # Dividing the array elements
        L = arr[:mid]

        # into 2 halves
        R = arr[mid:]

        # Sorting the first half
        mergeSort(L)

        # Sorting the second half
        mergeSort(R)

        i = j = k = 0

        # Copy data to temp arrays L[] and R[]
        while i < len(L) and j < len(R):
            if L[i] < R[j]:
                arr[k] = L[i]
                i += 1
            else:
                arr[k] = R[j]
                j += 1
            k += 1

        # Checking if any element was left
        while i < len(L):
            arr[k] = L[i]
            i += 1
            k += 1

        while j < len(R):
            arr[k] = R[j]
            j += 1
            k += 1


# -----------------------------------------------lucky number with two lucky any digits
res = set()


def solven(p, l, a, b, n):  # given number
    if p > n or l > 10:
        return
    if p > 0:
        res.add(p)
    solven(p * 10 + a, l + 1, a, b, n)
    solven(p * 10 + b, l + 1, a, b, n)


# problem
"""
n = int(input())
for a in range(0, 10):
    for b in range(0, a):
        solve(0, 0)
print(len(res))
"""


# Python3 program to find all subsets
# by backtracking.

# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index, d):
    print(*subset)
    s = sum(subset)
    d.append(s)
    for i in range(index, len(A)):
        # include the A[i] in subset.
        subset.append(A[i])

        # move onto the next element.
        subsetsUtil(A, subset, i + 1, d)

        # exclude the A[i] from subset and
        # triggers backtracking.
        subset.pop(-1)
    return d


def subsetSums(arr, l, r, d, sum=0):
    if l > r:
        d.append(sum)
        return
    subsetSums(arr, l + 1, r, d, sum + arr[l])

    # Subset excluding arr[l]
    subsetSums(arr, l + 1, r, d, sum)
    return d


def print_factors(x):
    factors = []
    for i in range(1, x + 1):
        if x % i == 0:
            factors.append(i)
    return (factors)


# -----------------------------------------------
def calc(X, d, ans, D):
    # print(X,d)
    if len(X) == 0:
        return

    i = X.index(max(X))
    ans[D[max(X)]] = d

    Y = X[:i]
    Z = X[i + 1:]

    calc(Y, d + 1, ans, D)
    calc(Z, d + 1, ans, D)


# ---------------------------------------

def factorization(n, l):
    c = n
    if prime(n) == True:
        l.append(n)
        return l
    for i in range(2, c):
        if n == 1:
            break
        while n % i == 0:
            l.append(i)
            n = n // i
    return l


# endregion------------------------------


def good(b):
    l = []
    i = 0
    while (len(b) != 0):
        if b[i] < b[len(b) - 1 - i]:
            l.append(b[i])
            b.remove(b[i])
        else:
            l.append(b[len(b) - 1 - i])
            b.remove(b[len(b) - 1 - i])
    if l == sorted(l):
        # print(l)
        return True
    return False


# arr=[]
# print(good(arr))
def generate(st, s):
    if len(s) == 0:
        return

    # If current string is not already present.
    if s not in st:
        st.add(s)

        # Traverse current string, one by one
        # remove every character and recur.
        for i in range(len(s)):
            t = list(s).copy()
            t.remove(s[i])
            t = ''.join(t)
            generate(st, t)
    return
#=--------------------------------------------longest increasing subsequence

def largestincreasingsubsequence(A):
    l = [1]*len(A)
    sub=[]
    for i in range(1,len(l)):
        for k in range(i):
            if A[k]<A[i]:
                sub.append(l[k])
        l[i]=1+max(sub,default=0)
    return max(l,default=0)

#----------------------------------longest palindromic substring




# Python3 program for the
# above approach

# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
    # Stores the prefix
    # sum of nums[]
    prefix_sum = 0

    # Stores the bitwise OR of
    # sum of each subsequence
    result = 0

    # Iterate through array nums[]
    for i in range(N):
        # Bits set in nums[i] are
        # also set in result
        result |= nums[i]

        # Calculate prefix_sum
        prefix_sum += nums[i]

        # Bits set in prefix_sum
        # are also set in result
        result |= prefix_sum

    # Return the result
    return result

#l=[]
def OR(a, n):
    ans = a[0]
    for i in range(1, n):
        ans |= a[i]
        #l.append(ans)
    return ans
#print(prime(12345678987766))
"""
def main():
    q=inpu()
    x = q
    v1 = 0
    v2 = 0
    i = 2
    while i * i <= q:
        while q % i == 0:
            if v1!=0:
                v2 = i
            else:
                v1 = i
            q //= i
        i += 1

    if q - 1!=0:
        v2 = q

    if v1 * v2 - x!=0:
        print(1)
        print(v1 * v2)
    else:
        print(2)
if __name__ == '__main__':
    main()
"""

"""
def main():
    l,r = sep()
    a=[]
    luckynum(0,l,r,a)
    a.sort()
    #print(a)
    i=0
    ans=0
    l-=1
    #print(a)
    while(True):
        if r>a[i]:
            ans+=(a[i]*(a[i]-l))
            l=a[i]
        else:
            ans+=(a[i]*(r-l))
            break
        i+=1
    print(ans)
if __name__ == '__main__':
    main()
"""
"""
def main():
    sqrt = {i * i: i for i in range(1, 1000)}
    #print(sqrt)
    a, b = sep()
    for y in range(1, a):
        x2 = a * a - y * y
        if x2 in sqrt:
            x = sqrt[x2]
            if b * y % a == 0 and b * x % a == 0 and b * x // a != y:
                print('YES')
                print(-x, y)
                print(0, 0)
                print(b * y // a, b * x // a)
                exit()
    print('NO')
if __name__ == '__main__':
    main()
"""
def main():
    m=inpu()
    q=lis()
    n=inpu()
    arr=lis()
    q=min(q)
    arr.sort(reverse=True)
    s=0
    cnt=0
    i=0
    while(i<n):
        cnt+=1
        s+=arr[i]
        #print(cnt,q)
        if cnt==q:
            i+=2
            cnt=0
        i+=1
    print(s)
if __name__ == '__main__':
    main()

```