#include <bits/stdc++.h>
using namespace std;
inline void readi(int &x);
const int N = 1000, M = 1000000007;
int C[N + 10][N + 10], fac[N + 10];
int f[N + 10][N + 10];
void am(int &a, const int &b) {
  a += b;
  if (a >= M) a -= M;
}
void init() {
  C[0][0] = 0;
  for (int i = 1; i <= N; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) am(C[i][j], C[i - 1][j - 1] + C[i - 1][j]);
  }
  fac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = (long long)fac[i - 1] * i % M;
  for (int l = 0; l <= N; l++) {
    f[l][1] = 1;
    for (int k = 2; k <= 65; k++) {
      int &t = f[l][k];
      for (int i = 0; (i + 1) * k - 1 <= l; i++) {
        t += f[l - (i + 1) * k + 1][k - 1];
        if (t >= M) t -= M;
      }
    }
  }
  for (int l = 0; l <= N; l++)
    for (int k = 1; k <= N; k++) f[l][k] = (long long)f[l][k] * fac[k] % M;
}
void work() {
  int T, n, k, ans;
  for (readi(T); T--;) {
    readi(n), readi(k);
    if (k > 60) {
      printf("0\n");
      continue;
    }
    ans = 0;
    for (int l = 0; l <= n; l++) am(ans, (long long)f[l][k] * C[n - l][k] % M);
    if (ans < 0) ans += M;
    printf("%d\n", ans);
  }
}
int main() {
  init();
  work();
  return 0;
}
inline void readi(int &x) {
  char c;
  for (c = getchar(); c > '9' || c < '0'; c = getchar())
    ;
  x = c ^ '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar())
    x = x * 10 + (c ^ '0');
}

### Prompt

Your challenge is to write 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;
inline void readi(int &x);
const int N = 1000, M = 1000000007;
int C[N + 10][N + 10], fac[N + 10];
int f[N + 10][N + 10];
void am(int &a, const int &b) {
  a += b;
  if (a >= M) a -= M;
}
void init() {
  C[0][0] = 0;
  for (int i = 1; i <= N; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) am(C[i][j], C[i - 1][j - 1] + C[i - 1][j]);
  }
  fac[0] = 1;
  for (int i = 1; i <= N; i++) fac[i] = (long long)fac[i - 1] * i % M;
  for (int l = 0; l <= N; l++) {
    f[l][1] = 1;
    for (int k = 2; k <= 65; k++) {
      int &t = f[l][k];
      for (int i = 0; (i + 1) * k - 1 <= l; i++) {
        t += f[l - (i + 1) * k + 1][k - 1];
        if (t >= M) t -= M;
      }
    }
  }
  for (int l = 0; l <= N; l++)
    for (int k = 1; k <= N; k++) f[l][k] = (long long)f[l][k] * fac[k] % M;
}
void work() {
  int T, n, k, ans;
  for (readi(T); T--;) {
    readi(n), readi(k);
    if (k > 60) {
      printf("0\n");
      continue;
    }
    ans = 0;
    for (int l = 0; l <= n; l++) am(ans, (long long)f[l][k] * C[n - l][k] % M);
    if (ans < 0) ans += M;
    printf("%d\n", ans);
  }
}
int main() {
  init();
  work();
  return 0;
}
inline void readi(int &x) {
  char c;
  for (c = getchar(); c > '9' || c < '0'; c = getchar())
    ;
  x = c ^ '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar())
    x = x * 10 + (c ^ '0');
}

```