#include <bits/stdc++.h>
using namespace std;
int c[2005][2005] = {0};
int fac[5005];
int f[5005][100] = {0};
void inic() {
  c[0][0] = 1;
  for (int i = 1; i <= 1102; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
  }
}
int A(int a, int b) {
  if (a > b) return 0;
  return c[b - 1][a - 1];
}
int B(int a, int b) { return 1ll * f[b][a] * fac[a] % 1000000007; }
void inif() {
  f[0][0] = 1;
  for (int i = 1; i <= 1100; i++)
    for (int k = 1; k <= 70; k++) {
      int an = 0;
      for (int j = 1; j * k <= i; j++)
        an = (an + f[i - j * k][k - 1]) % 1000000007;
      f[i][k] = an;
    }
}
int main() {
  int tes;
  inic();
  inif();
  fac[0] = 1;
  for (int i = 1; i <= 3002; i++) fac[i] = 1ll * fac[i - 1] * i % 1000000007;
  scanf("%d", &tes);
  while (tes--) {
    int k, n;
    scanf("%d%d", &n, &k);
    if (k > 50)
      printf("0\n");
    else {
      int ans = 0;
      for (int i = 0; i <= n + 1; i++) {
        ans = (ans + 1ll * A(k + 1, i) * B(k, n + 1 - i + k)) % 1000000007;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
int c[2005][2005] = {0};
int fac[5005];
int f[5005][100] = {0};
void inic() {
  c[0][0] = 1;
  for (int i = 1; i <= 1102; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
  }
}
int A(int a, int b) {
  if (a > b) return 0;
  return c[b - 1][a - 1];
}
int B(int a, int b) { return 1ll * f[b][a] * fac[a] % 1000000007; }
void inif() {
  f[0][0] = 1;
  for (int i = 1; i <= 1100; i++)
    for (int k = 1; k <= 70; k++) {
      int an = 0;
      for (int j = 1; j * k <= i; j++)
        an = (an + f[i - j * k][k - 1]) % 1000000007;
      f[i][k] = an;
    }
}
int main() {
  int tes;
  inic();
  inif();
  fac[0] = 1;
  for (int i = 1; i <= 3002; i++) fac[i] = 1ll * fac[i - 1] * i % 1000000007;
  scanf("%d", &tes);
  while (tes--) {
    int k, n;
    scanf("%d%d", &n, &k);
    if (k > 50)
      printf("0\n");
    else {
      int ans = 0;
      for (int i = 0; i <= n + 1; i++) {
        ans = (ans + 1ll * A(k + 1, i) * B(k, n + 1 - i + k)) % 1000000007;
      }
      printf("%d\n", ans);
    }
  }
  return 0;
}

```