"""
Given an array A of N positive numbers. The task is to find the position where equilibrium first occurs in the array.

Equilibrium position in an array is a position such that the sum of elements before it is equal to the sum of elements after it.

Input:
The first line of input contains an integer T, denoting the number of test cases. Then T test cases follow.
First line of each test case contains an integer N denoting the size of the array. 
Then in the next line are N space separated values of the array A.

Output:
For each test case in a new  line print the position at which the elements are at equilibrium if no equilibrium point exists print -1.

Constraints:
1 <= T <= 100
1 <= N <= 106
1 <= Ai <= 108

Example:
Input:
2
1
1
5
1 3 5 2 2

Output:
1
3

Explanation:
Testcase 1: Since its the only element hence its the only equilibrium point.
Testcase 2: For second test case equilibrium point is at position 3 as elements below it (1+3) = elements after it (2+2).
"""
T= int(input())
for i in range(T):
    N=int(input())
    A=list(map(int,input().split()))
    s=sum(A)
    r=s-A[0]
    l=0
    k=-1
    if(r==l):
        print(1)
        continue
    for i in range(1,N):
        if(r==l):
            k=i
            break
        r=r-A[i]
        l=l+A[i-1]
    print(k)    

2b7d5dea048469d6865fe28e3489313c190f4a94
<mask token>

<mask token>
for i in range(T):
    N = int(input())
    A = list(map(int, input().split()))
    s = sum(A)
    r = s - A[0]
    l = 0
    k = -1
    if r == l:
        print(1)
        continue
    for i in range(1, N):
        if r == l:
            k = i
            break
        r = r - A[i]
        l = l + A[i - 1]
    print(k)

<mask token>
T = int(input())
for i in range(T):
    N = int(input())
    A = list(map(int, input().split()))
    s = sum(A)
    r = s - A[0]
    l = 0
    k = -1
    if r == l:
        print(1)
        continue
    for i in range(1, N):
        if r == l:
            k = i
            break
        r = r - A[i]
        l = l + A[i - 1]
    print(k)

"""
Given an array A of N positive numbers. The task is to find the position where equilibrium first occurs in the array.

Equilibrium position in an array is a position such that the sum of elements before it is equal to the sum of elements after it.

Input:
The first line of input contains an integer T, denoting the number of test cases. Then T test cases follow.
First line of each test case contains an integer N denoting the size of the array. 
Then in the next line are N space separated values of the array A.

Output:
For each test case in a new  line print the position at which the elements are at equilibrium if no equilibrium point exists print -1.

Constraints:
1 <= T <= 100
1 <= N <= 106
1 <= Ai <= 108

Example:
Input:
2
1
1
5
1 3 5 2 2

Output:
1
3

Explanation:
Testcase 1: Since its the only element hence its the only equilibrium point.
Testcase 2: For second test case equilibrium point is at position 3 as elements below it (1+3) = elements after it (2+2).
"""
T= int(input())
for i in range(T):
    N=int(input())
    A=list(map(int,input().split()))
    s=sum(A)
    r=s-A[0]
    l=0
    k=-1
    if(r==l):
        print(1)
        continue
    for i in range(1,N):
        if(r==l):
            k=i
            break
        r=r-A[i]
        l=l+A[i-1]
    print(k)    
