#include <bits/stdc+

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

bool check(ll u, ll v, ll w, ll x) {
if (u * x + v == w) return true;
if (u * x + w == v) return true;
if (v * x + u == w) return true;
if (v * x + w == u) return true;
if (w * x + u == v) return true;
if (w * x + v == u) return true;
return false;
}

vector<ll> get_candidates(ll u, ll v, ll w) {
vector<ll> res;
if ((w - v) % u == 0) {
ll x = (w - v) / u;
if (x >= 0) res.push_back(x);
}
if ((v - w) % u == 0) {
ll x = (v - w) / u;
if (x >= 0) res.push_back(x);
}
if ((w - u) % v == 0) {
ll x = (w - u) / v;
if (x >= 0) res.push_back(x);
}
if ((u - w) % v == 0) {
ll x = (u - w) / v;
if (x >= 0) res.push_back(x);
}
if ((v - u) % w == 0) {
ll x = (v - u) / w;
if (x >= 0) res.push_back(x);
}
if ((u - v) % w == 0) {
ll x = (u - v) / w;
if (x >= 0) res.push_back(x);
}
sort(res.begin(), res.end());
res.erase(unique(res.begin(), res.end()), res.end());
return res;
}

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

int T;
cin >> T;
while (T--) {
    int n;
    cin >> n;
    vector<array<ll, 3>> eqs(n);
    for (int i = 0; i < n; i++) {
        cin >> eqs[i][0] >> eqs[i][1] >> eqs[i][2];
    }

    // 从第一条方程获取候选 x —— 修正了这一行！
    vector<ll> candidates = get_candidates(eqs[0], eqs[1], eqs[2](@ref);  // 修正：eqs[0] 即 u, eqs[1] 即 v, eqs[2] 即 w

    ll ans = -1;
    for (ll x : candidates) {
        bool ok = true;
        for (int i = 0; i < n && ok; i++) {
            if (!check(eqs[i][0], eqs[i][1], eqs[i][2], x)) {
                ok = false;
            }
        }
        if (ok) {
            ans = x;
            break;
        }
    }

    cout << ans << '\n';
}
return 0;

}