The sequence of integer pairs (a1, b1), (a2, b2), ..., (ak, bk) is beautiful, if the following statements are fulfilled:

• 1 ≤ a1 ≤ b1 < a2 ≤ b2 < ... < ak ≤ bk ≤ n, where n is a given positive integer;
• all numbers b1 - a1, b2 - a2, ..., bk - ak are distinct.

For the given number n find the number of beautiful sequences of length k. As the answer can be rather large, print the remainder after dividing it by 1000000007 (109 + 7).

The first line contains integer t (1 ≤ t ≤  2·105) — the number of the test data.

Each of the next t lines contains two integers n and k (1 ≤ k ≤ n ≤ 1000).

For each test from the input print the answer to the problem modulo 1000000007 (109 + 7). Print the answers to the tests in the order in which the tests are given in the input.

In the first test sample there is exactly one beautiful sequence: (1, 1).

In the second test sample, the following sequences are beautiful:

• (1, 1);
• (1, 2);
• (2, 2).

In the fourth test sample, the following sequences are beautiful:

• (1, 1);
• (1, 2);
• (1, 3);
• (2, 2);
• (2, 3);
• (3, 3).

In the fifth test sample, the following sequences are beautiful:

• (1, 1), (2, 3);
• (1, 2), (3, 3).

In the third and sixth samples, there are no beautiful sequences.

dp 背包问题

预处理fac[i] 为 i!   c[i][j] 为组合数 dp[i][j] 代表 i 体积中装 j 个物品

dp[i][j] = dp[i - j][j - 1] + dp[i - j][j];

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 using namespace std;

 typedef long long ll;

 const int MOD = 1e9 + ,N = ;

 ll c[N + ][N + ],dp[N + ][N + ],fac[N + ];

 void init() {

         fac[] = ;

         for(int i = ; i <= N; ++i) {

                 fac[i] = fac[i - ] * i % MOD;

         }

         for(int i = ; i <= N; ++i) {

                 c[i][] = c[i][i] = ;

                 for(int j = ; j < i; ++j) {

                         c[i][j] = (c[i - ][j] + c[i - ][j - ]) % MOD;

                 }

         }

         dp[][] = ;

         for(int i = ; i <= N; ++i) {

                 for(int j = ; j <= i; ++j) {

                         dp[i][j] = (dp[i - j][j] + dp[i - j][j - ]) % MOD;

                 }

         }

 }

 int main()

 {

     init();

     int t;

     scanf("%d",&t);

     while(t--) {

             int n,k,ans = ;

             scanf("%d%d",&n,&k);

             for(int s = k * (k + ) / ; s <= n; ++s) {

                     ans = (ans + dp[s][k] * c[n - s + k][k]) % MOD;

             }

             printf("%d\n",ans * fac[k] % MOD);

     }

     //cout << "Hello world!" << endl;

     return ;

 }