#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1000000007, N = 1010, D = 55, M = 2010;
long long t;
long long p[N][N], s[N][N], bc[M][M];
long long mod(long long x) {
  x %= MOD;
  if (x < 0) x += MOD;
  return x;
}
void init() {
  for (long long i = 0; i < M; i++) bc[i][0] = 1;
  for (long long i = 1; i < M; i++)
    for (long long j = 1; j <= i; j++)
      bc[i][j] = mod(bc[i - 1][j - 1] + bc[i - 1][j]);
  p[0][0] = 1;
  for (long long k = 1; k < N; k++)
    for (long long i = N - 1; i; i--)
      for (long long j = D; j; j--)
        if (i >= k) p[i][j] = mod(p[i][j] + p[i - k][j - 1]);
  for (long long i = 1; i < N; i++)
    for (long long j = 1; j < D; j++) {
      long long s = 1;
      for (long long k = 1; k <= j; k++) s = mod(s * k);
      p[i][j] = mod(p[i][j] * s);
    }
  for (long long i = 1; i < N; i++)
    for (long long j = 1; j < D; j++)
      for (long long l = 1; l <= i; l++)
        s[i][j] = mod(s[i][j] + mod(p[l][j] * bc[i - l + j][j]));
}
int main() {
  init();
  scanf("%I64d", &t);
  while (t--) {
    long long x, y;
    scanf("%I64d %I64d", &x, &y);
    printf("%I64d\n", s[x][y]);
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 long long MOD = 1000000007, N = 1010, D = 55, M = 2010;
long long t;
long long p[N][N], s[N][N], bc[M][M];
long long mod(long long x) {
  x %= MOD;
  if (x < 0) x += MOD;
  return x;
}
void init() {
  for (long long i = 0; i < M; i++) bc[i][0] = 1;
  for (long long i = 1; i < M; i++)
    for (long long j = 1; j <= i; j++)
      bc[i][j] = mod(bc[i - 1][j - 1] + bc[i - 1][j]);
  p[0][0] = 1;
  for (long long k = 1; k < N; k++)
    for (long long i = N - 1; i; i--)
      for (long long j = D; j; j--)
        if (i >= k) p[i][j] = mod(p[i][j] + p[i - k][j - 1]);
  for (long long i = 1; i < N; i++)
    for (long long j = 1; j < D; j++) {
      long long s = 1;
      for (long long k = 1; k <= j; k++) s = mod(s * k);
      p[i][j] = mod(p[i][j] * s);
    }
  for (long long i = 1; i < N; i++)
    for (long long j = 1; j < D; j++)
      for (long long l = 1; l <= i; l++)
        s[i][j] = mod(s[i][j] + mod(p[l][j] * bc[i - l + j][j]));
}
int main() {
  init();
  scanf("%I64d", &t);
  while (t--) {
    long long x, y;
    scanf("%I64d %I64d", &x, &y);
    printf("%I64d\n", s[x][y]);
  }
  return 0;
}

```