/*

Given an array having both positive and negative integers. The task is to compute the length of the largest subarray with sum 0.

Input:
The first line of input contains an element T denoting the number of test cases. Then T test cases follow. Each test case consists of 2 lines. The first line of each test case contains a number denoting the size of the array A. Then in the next line are space-separated values of the array A.

Output:
For each test case, the output will be the length of the largest subarray which has sum 0.

User Task:
Since this is a functional problem you don't have to worry about input, you just have to complete the function maxLen() which takes two arguments an array A and n, where n is the size of the array A and returns the length of the largest subarray with 0 sum.

Expected Time Complexity: O(N*Log(N)).
Expected Auxiliary Space: O(N).

Constraints:
1 <= T <= 100
1 <= N <= 104
-1000 <= A[i] <= 1000, for each valid i

Example:
Input
1
8
15 -2 2 -8 1 7 10 23

Output
5

Explanation
Testcase 1: In the above test case the largest subarray with sum 0 will be -2 2 -8 1 7.
*/











#include <bits/stdc++.h>
using namespace std;

int maxLen(int A[], int n);

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int N;
        cin >> N;
        int A[N];
        for (int i = 0; i < N; i++)
            cin >> A[i];
        cout << maxLen(A, N) << endl;
    }
}




int maxLen(int a[], int n)
{
   int maxLen=0;
   unordered_map<int, int> m;
   int sum=0;
   for(int i=0;i<n;i++){
       sum+=a[i];
       if(sum==0) maxLen=max(maxLen, i+1);
       if(m.find(sum)!=m.end()) maxLen=max(maxLen, i-m[sum]);
       else m.insert({sum, i});
   }
   return maxLen;
}