p03413 AtCoder Regular Contest 092 - Both Sides Merger
You have an integer sequence of length N: a_1, a_2, ..., a_N.

You repeatedly perform the following operation until the length of the sequence becomes 1:

* First, choose an element of the sequence.
* If that element is at either end of the sequence, delete the element.
* If that element is not at either end of the sequence, replace the element with the sum of the two elements that are adjacent to it. Then, delete those two elements.



You would like to maximize the final element that remains in the sequence.

Find the maximum possible value of the final element, and the way to achieve it.

Constraints

* All input values are integers.
* 2 \leq N \leq 1000
* |a_i| \leq 10^9

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

* In the first line, print the maximum possible value of the final element in the sequence.
* In the second line, print the number of operations that you perform.
* In the (2+i)-th line, if the element chosen in the i-th operation is the x-th element from the left in the sequence at that moment, print x.
* If there are multiple ways to achieve the maximum value of the final element, any of them may be printed.

Examples

Input

5
1 4 3 7 5


Output

11
3
1
4
2


Input

4
100 100 -1 100


Output

200
2
3
1


Input

6
-1 -2 -3 1 2 3


Output

4
3
2
1
2


Input

9
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000


Output

5000000000
4
2
2
2
2
```python
def examC():
    ans = 0
    print(ans)
    return

def examD():
    ans = 0
    print(ans)
    return

def examE():
    N = I()
    A = LI()
    ans = -inf

    S = [0]*(N+1)
    fr = [-1]*N
    best = -1

    for i,a in enumerate(A):
        S[i] = a
        for j in range(i):
            if ((i-j)%2==0):
                if (S[j]+a>S[i]):
                    fr[i] = j
                    S[i] = S[j] + a
            if (S[i]>ans):
                ans = S[i]
                best = i
    #print(best)
    V = []
    for i in range(best+1,N)[::-1]:
        V.append(i+1)
    i = best
    while(fr[i]>=0):
        f = fr[i]
        #print(i,f)
        while(f<i):
            V.append(1+(i+f)//2)
            i -= 2
    for _ in range(i):
        V.append(1)

    print(ans)
    print(len(V))
    for v in V:
        print(v)
    return

def examF():
    ans = 0
    print(ans)
    return

from decimal import getcontext,Decimal as dec
import sys,bisect,itertools,heapq,math,random
from copy import deepcopy
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
def I(): return int(input())
def LI(): return list(map(int,sys.stdin.readline().split()))
def DI(): return dec(input())
def LDI(): return list(map(dec,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def SI(): return sys.stdin.readline().strip()
global mod,mod2,inf,alphabet,_ep
mod = 10**9 + 7
mod2 = 998244353
inf = 10**18
_ep = dec("0.000000000001")
alphabet = [chr(ord('a') + i) for i in range(26)]
alphabet_convert = {chr(ord('a') + i): i for i in range(26)}

getcontext().prec = 28

sys.setrecursionlimit(10**7)

if __name__ == '__main__':
    examE()

"""
142
12 9 1445 0 1
asd dfg hj o o
aidn
"""
```