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

using int64 = long long;

int64 count_two_robots_multiset(const vector<int64>& w) {
    int n = (int)w.size();

    vector<int64> pref(n + 1, 0);
    for (int i = 0; i < n; i++) {
        pref[i + 1] = pref[i] + w[i];
    }

    // The input guarantees total sum is zero.
    assert(pref[n] == 0);

    // Rotate so that all prefixes are non-negative.
    // This makes the circular structure linear.
    int cut = 0;
    for (int i = 0; i < n; i++) {
        if (pref[i] < pref[cut]) {
            cut = i;
        }
    }

    vector<int64> a(n);
    for (int i = 0; i < n; i++) {
        a[i] = w[(cut + i) % n];
    }

    // P[i] = signed prefix before station i in the rotated array.
    // D[i] = total sink demand before station i in the rotated array.
    vector<int64> P(n + 1, 0), D(n + 1, 0);
    for (int i = 0; i < n; i++) {
        P[i + 1] = P[i] + a[i];
        D[i + 1] = D[i] + max<int64>(0, -a[i]);
    }

    int64 ans = 0;

    // Count pairs of two distinct starting positions.
    // In the rotated order, count pairs (i, j) with i < j.
    int j = 1;
    for (int i = 0; i < n; i++) {
        j = max(j, i + 1);

        int64 need = D[i] + P[i];
        while (j < n && D[j] < need) {
            j++;
        }

        // Every j, j + 1, ..., n - 1 works with i.
        ans += n - j;
    }

    // Count pairs where both robots start at the same position.
    // This is possible exactly when a single robot starting there can solve everything,
    // i.e. the corresponding prefix is a global minimum. After this rotation, those are
    // exactly the positions with P[i] == 0.
    for (int i = 0; i < n; i++) {
        if (P[i] == 0) {
            ans++;
        }
    }

    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int t;
    cin >> t;

    while (t--) {
        int n;
        cin >> n;

        vector<int64> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }

        cout << count_two_robots_multiset(a) << '\n';
    }

    return 0;
}