The Chef is given an array of numbers and asked to find the first balance point of an array. A balance point is an index in the array N where the sum of the integers to the left of N is equal to the sum of the integers to the right of N. If there is no such balance point, print -1
-----Input-----
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains a single integer N denoting the number of weights on the array. The second line contains N space-separated integers A1, A2, ..., AN denoting the weights
-----Output-----
For each test case, output a single line containing the index of the first balance point. If no such index exists, output -1
-----Constraints-----
For Subtask 1:
- 1 ≤ T ≤ 10
- 1 ≤ N ≤ 10
- -1000 ≤ A1, A2, ..., AN ≤ 1000
For Subtask 1:
- 1 ≤ T ≤ 10
- 1 ≤ N ≤ 1000
- -10000 ≤ A1, A2, ..., AN ≤ 10000
-----Example-----
Input:
2
7
7 8 9 1 9 8 7
6
1 100 50 -51 1 1
Output:
3
1
-----Explanation-----
Example case 1. In the first case, at the midpoint of the array 1, the sum of the left side {7,8,9} is equal to the sum of the right side {9,8,7}
Example case 1. In the second case, at index 1, the sum of left side =1 is equal to the sum of the right side {50,-51,1,1}

l=[]
n=int(input())
for i in range(0,n):
 s=int(input())
 l=[]
 l=list(map(int,input().split()))
 beg=0
 suml=0
 sumr=0
 for j in range(0,len(l)):
  sumr=sumr+l[j]
 for j in range(0,len(l)):
  sumr=sumr-l[j]
  if(sumr==suml):
   print(j)
   beg=1
   break
  suml=suml+l[j]
 if(beg==0):
  print(-1)

l=[]
n=int(input())
for i in range(0,n):
 s=int(input())
 l=[]
 l=list(map(int,input().split()))
 beg=0
 suml=0
 sumr=0
 for j in range(0,len(l)):
  sumr=sumr+l[j]
 for j in range(0,len(l)):
  sumr=sumr-l[j]
  if(sumr==suml):
   print(j)
   beg=1
   break
  suml=suml+l[j]
 if(beg==0):
  print(-1)