#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MOD = 1000000000 + 7;
const int MAXN = 1000000; // MAXM=MAXN
int fact[MAXN + 5];
int factInv[MAXN + 5]; // (i!)^{-1}
// n 个元素错位排列数
// n! * SUM{k=0~n}((-1)^k/k!)
int d[MAXN + 5];
// a^b (mod p)
long long quick_pow(long long a, long long b, long long p)
{
    long long res = 1;
    while (b)
    {
        if (b & 1)
            res = (res * a) % p;
        b = b >> 1;
        a = (a * a) % p;
    }
    return res;
}
// 求 x 在模 p 意义下的逆元
int inv(int x, int p)
{
    return quick_pow(x, p - 2, p);
}
// 计算fact、inv、d
void init()
{
    fact[0] = 1;
    for (int i = 1; i <= MAXN; i++)
        fact[i] = fact[i - 1] * i % MOD;
    for (int i = 0; i <= MAXN; i++)
        factInv[i] = inv(fact[i], MOD);
    d[0] = 1;
    d[1] = 0;
    for (int i = 2; i <= 1000000; i++)
        d[i] = (i - 1) * (d[i - 1] + d[i - 2]) % MOD;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    init();
    while (T--)
    {
        int n, m;
        cin >> n >> m;
        // 选m个位置固定 C(n,m) = n!/(m!(n-m)!)
        // 剩下的 n-m 个位置错排 d[n-m]
        int ans = 1;
        ans = (ans * fact[n]) % MOD;
        ans = (ans * factInv[m]) % MOD;
        ans = (ans * factInv[n - m]) % MOD;
        ans = (ans * d[n - m]) % MOD;
        cout << ans << "\n";
    }
    return 0;
}