#include <bits/stdc++.h>
#include <atcoder/all>
typedef long long int ll;
typedef long long int ull;
#define MP make_pair
using namespace std;
using namespace atcoder;
typedef pair<ll, ll> P;
// const ll MOD = 998244353;
const ll MOD = 1000000007;
// using mint = modint998244353;
using mint = modint1000000007;
const double pi = 3.1415926536;
const int MAX = 2000003;
long long fac[MAX], finv[MAX], inv[MAX];
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll gcd(ll x, ll y) {
   if (y == 0) return x;
   else if (y > x) {
       return gcd (y, x); 
   }
   else return gcd(x % y, y);
}
ll lcm(ll x, ll y) {
   return x / gcd(x, y) * y;
}
ll my_sqrt(ll x) {
    ll m = 0;
    ll M = 3000000001;
    while (M - m > 1) {
        ll now = (M + m) / 2;
        if (now * now <= x) {
            m = now;
        }
        else {
            M = now;
        }
    }
    return m;
}
ll keta(ll n) {
    ll ret = 0;
    while (n) {
        n /= 10;
        ret++;
    }
    return ret;
}
ll ceil(ll n, ll m) {
    // n > 0, m > 0
    ll ret = n / m;
    if (n % m) ret++;
    return ret;
}
ll pow_ll(ll x, ll n) {
    if (n == 0) return 1;
    if (n % 2) {
        return pow_ll(x, n - 1) * x;
    }
    else {
        ll tmp = pow_ll(x, n / 2);
        return tmp * tmp;
    }
}

int main() {
    int q;
    cin >> q;
    for (int t = 0; t < q; t++) {
        ll n, i, j;
        cin >> n >> i >> j;
        ll k = min({i, j, n - 1 - i, n - 1 - j});
        ll ans = 4 * k * (n - k);
        n -= 2 * k;
        i -= k;
        j -= k;
        // cout << ans << endl;
        if (i == 0) {
            cout << ans + j << endl;
        }
        else if (j == n - 1) {
            cout << ans + n - 1 + i << endl;
        }
        else if (i == n - 1) {
            cout << ans + 2 * (n - 1) + (n - 1 - j);
        }
        else {
            cout << ans + 3 * (n - 1) + (n - 1 - i);
        }
    }
    return 0;
}