1557_A. Ezzat and Two Subsequences_175
Ezzat has an array of n integers (maybe negative). He wants to split it into two non-empty subsequences a and b, such that every element from the array belongs to exactly one subsequence, and the value of f(a) + f(b) is the maximum possible value, where f(x) is the average of the subsequence x. 

A sequence x is a subsequence of a sequence y if x can be obtained from y by deletion of several (possibly, zero or all) elements.

The average of a subsequence is the sum of the numbers of this subsequence divided by the size of the subsequence.

For example, the average of [1,5,6] is (1+5+6)/3 = 12/3 = 4, so f([1,5,6]) = 4.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^3)— the number of test cases. Each test case consists of two lines.

The first line contains a single integer n (2 ≤ n ≤ 10^5).

The second line contains n integers a_1, a_2, …, a_n (-10^9 ≤ a_i ≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 3⋅10^5.

Output

For each test case, print a single value — the maximum value that Ezzat can achieve.

Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Example

Input


4
3
3 1 2
3
-7 -6 -6
3
2 2 2
4
17 3 5 -3


Output


4.500000000
-12.500000000
4.000000000
18.666666667

Note

In the first test case, the array is [3, 1, 2]. These are all the possible ways to split this array: 

  * a = [3], b = [1,2], so the value of f(a) + f(b) = 3 + 1.5 = 4.5. 
  * a = [3,1], b = [2], so the value of f(a) + f(b) = 2 + 2 = 4. 
  * a = [3,2], b = [1], so the value of f(a) + f(b) = 2.5 + 1 = 3.5. 

Therefore, the maximum possible value 4.5.

In the second test case, the array is [-7, -6, -6]. These are all the possible ways to split this array: 

  * a = [-7], b = [-6,-6], so the value of f(a) + f(b) = (-7) + (-6) = -13. 
  * a = [-7,-6], b = [-6], so the value of f(a) + f(b) = (-6.5) + (-6) = -12.5. 

Therefore, the maximum possible value -12.5.
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat, permutations
import math
import heapq

MOD = 998244353
def inp(force_list=False):
    re = map(int, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inpf(force_list=False):
    re = map(float, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inst():
    return raw_input().strip()

def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x

def lcm(x, y):
    return x*y/(gcd(x,y))

def qmod(a, b, mod=MOD):
    res = 1
    while b:
        if b&1:
            res = (res*a)%mod
        b >>= 1
        a = (a*a)%mod
        # print b
    return res

def inv(a):
    return qmod(a, MOD-2)

def query(num, k):
    print tmk(num, k)
    stdout.flush()
    return inp()

class RMQ(object):
    def __init__(self, da):
        self.n = len(da)
        self.da = [[0 for i in range(32)] for i in range(self.n)]
        for i in range(self.n):
            self.da[i][0] = da[i]
        for j in range(1, 32):
            for i in range(self.n):
                self.da[i][j] = gcd(self.da[i][j-1], self.da[(i+(1<<(j-1)))%self.n][j-1])

    def query(self, l, r):
        j = int(math.log(r-l+1, 2))
        return gcd(self.da[l][j], self.da[(r-(1<<j)+1)%self.n][j])


def getsum(BITTree,i):
    if i<0:
        return 0
    s = 0
    i = i+1
    while i > 0:
        s += BITTree[i]
        i -= i & (-i)
    return s

def updatebit(BITTree , n , i ,v):
    i += 1
    while i <= n:
        BITTree[i] += v
        i += i & (-i)

def construct(arr, n):

    BITTree = [0]*(n+1)
    for i in range(n):
        updatebit(BITTree, n, i, arr[i])
    return BITTree

def GV(num, s, p):
    num = num**s - (num%p)**s
    re = 0
    while num and num%p==0:
        re+=1
        num/=p
    return re

def my_main():
    kase = inp()
    pans = []
    for ka in range(1, kase+1):
        # print "Case #%d:" % ka,
        n = inp()
        da = inp()
        da.sort()
        pans.append(str( da[-1]/float(1) + sum(da[:-1])/float(n-1)))
    print '\n'.join(pans)


my_main()
