组合数学问题：棋盘上的组合数计算

#include 
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;

// 扩展欧几里得算法求解乘法逆元
ll extended_gcd(ll a, ll b, ll &x, ll &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    ll ans = extended_gcd(b, a % b, x, y);
    ll temp = x;
    x = y;
    y = temp - a / b * y;
    return ans;
}

// 计算乘法逆元
ll mod_inverse(ll a, ll n) {
    ll x, y;
    ll d = extended_gcd(a, n, x, y);
    if (d == 1)
        return (x % n + n) % n;
    else
        return -1;
}

// 计算组合数C(n, m)
ll combination(ll n, ll m) {
    ll ans = 1;
    for (ll i = n; i >= n - m + 1; i--) {
        ans = (ans * i) % MOD;
    }
    for (ll i = 1; i <= m; i++) {
        ans = (ans * mod_inverse(i, MOD)) % MOD;
    }
    return ans;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n, m;
        cin >> n >> m;
        if (n         cout <    }
    return 0;
}