#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7, M = 1000, K = 45;
int n, m;
int fac[50];
int dp[2][50][M + 5], sum[M + 5][50][M + 5];
void add(int &a, int b) {
  a += b;
  if (a >= P) a -= P;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i <= K; i++) fac[i] = 1ll * fac[i - 1] * i % P;
  dp[0][0][0] = 1;
  for (int i = 0; i <= M; i++) sum[0][0][i] = 1;
  int cur = 0;
  for (int i = 1; i <= M; i++) {
    cur = 1 - cur;
    memset(dp[cur], 0, sizeof(dp[cur]));
    for (int j = 0; j <= K && (1 + j) * j / 2 <= i; j++) {
      for (int k = 0; k <= M; k++) {
        dp[cur][j][k] = dp[1 - cur][j][k];
        if (j > 0 && i >= k) {
          add(dp[cur][j][k], sum[i - k][j - 1][k - 1]);
        }
        sum[i][j][k] = sum[i][j][k - 1];
        add(sum[i][j][k], dp[cur][j][k]);
      }
    }
  }
}
int main() {
  int T;
  init();
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &m);
    if (m * (m + 1) / 2 > n)
      printf("0\n");
    else {
      printf("%d\n", 1ll * sum[n][m][M] * fac[m] % P);
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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).

Input

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).

Output

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.

Examples

Input

6
1 1
2 1
2 2
3 1
3 2
3 3


Output

1
3
0
6
2
0

Note

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7, M = 1000, K = 45;
int n, m;
int fac[50];
int dp[2][50][M + 5], sum[M + 5][50][M + 5];
void add(int &a, int b) {
  a += b;
  if (a >= P) a -= P;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i <= K; i++) fac[i] = 1ll * fac[i - 1] * i % P;
  dp[0][0][0] = 1;
  for (int i = 0; i <= M; i++) sum[0][0][i] = 1;
  int cur = 0;
  for (int i = 1; i <= M; i++) {
    cur = 1 - cur;
    memset(dp[cur], 0, sizeof(dp[cur]));
    for (int j = 0; j <= K && (1 + j) * j / 2 <= i; j++) {
      for (int k = 0; k <= M; k++) {
        dp[cur][j][k] = dp[1 - cur][j][k];
        if (j > 0 && i >= k) {
          add(dp[cur][j][k], sum[i - k][j - 1][k - 1]);
        }
        sum[i][j][k] = sum[i][j][k - 1];
        add(sum[i][j][k], dp[cur][j][k]);
      }
    }
  }
}
int main() {
  int T;
  init();
  scanf("%d", &T);
  while (T--) {
    scanf("%d %d", &n, &m);
    if (m * (m + 1) / 2 > n)
      printf("0\n");
    else {
      printf("%d\n", 1ll * sum[n][m][M] * fac[m] % P);
    }
  }
  return 0;
}

```