#include <bits/stdc++.h>
using namespace std;
#define int long long int
#define double long double
inline int power(int a, int b) {
int x = 1;
while (b) {
if (b & 1) x *= a;
a *= a;
b >>= 1;
}
return x;
}
const int M = 1000000007;
const int N = 3e5+9;
const int INF = 2e9+1;
const int LINF = 2000000000000000001;
//_ ***************************** START Below *******************************
// In Good Subarrays problem, draw array && subarray and apply Maths and find condition
// L R
// [ _ _ _ _ _ (_ _ _ _ _ _) _ _ ]
// Given : Px[R] - Px[L-1] == Py[R] - Py[L-1]
// cnt of x == cnt of y in subarray [L,R]
// Px[R]-Py[R] == Px[L-1] - Py[L-1]
// Hence find for Px[R] - Py[R] in past
vector<int> a;
void consistency(int n) {
int ones = 0;
int zeroes = 0;
unordered_map<int, int> mp = {{0, -1}};
int ans = 0;
for(int i=0; i<n; i++){
if(a[i] == 0) zeroes++;
else ones++;
if(mp.count(zeroes-ones)){
int l = mp[zeroes-ones];
int len = i-l;
ans = max(ans, len);
}
if(!mp.count(zeroes-ones)) mp[zeroes-ones] = i;
}
cout << ans << endl;
}
void solve() {
int n;
cin >> n;
a.resize(n);
for(int i=0; i<n; i++) cin >> a[i];
consistency(n) ;
}
int32_t main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}
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