Problem Description

Input

Output

For each test case, output the answer mod 1000000007(10^9 + 7).

Sample Input

2 1 1 3 1

Sample Output

1 4

Source

第八届福建省大学生程序设计竞赛-重现赛（感谢承办方厦门理工学院）

#define _CRT_SECURE_NO_DEPRECATE
#include <iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<bitset>
#include<set>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
#define N_MAX 10000+4
#define MOD 1000000007
#define INF 0x3f3f3f3f
typedef long long ll;
int n, k;
ll dp[N_MAX];//错排数
ll C[N_MAX][100 + 2];
void init() {
    dp[0] = 1; dp[1] = 0; dp[2] = 1;
    for (int i = 3; i < N_MAX; i++) {
        dp[i] = ((i - 1)*(dp[i - 1] + dp[i - 2]) + MOD) % MOD;
    }
}
void C_table() {
    for (int i = 0; i < N_MAX; i++) {
        C[i][0] = 1;
        for (int j = 1; j <= min(100, i); j++) {//!!!j<=i并且数组中j不能超过100
            C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
        }
    }
}

int main() {
    init(); C_table();
    int t; scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &k);
        ll num = 0, ans = 1;
        for (ll i = 1; i <= n; i++) {
            ans = ans*i%MOD;
        }
        for (int i = 0; i <= k - 1; i++) {
            num = (num + C[n][i] * dp[n - i]) % MOD;
        }
        printf("%lld\n", (ans - num + MOD) % MOD);
    }
    return 0;
}