#include <bits/stdc++.h>
using namespace std;
int dp[50][1010][1010];
long long fc[1010];
long long c1[1010][1010];
long long pw(long long x, long long p) {
  if (p == 0) return 1;
  long long z = pw(x, p / 2);
  z *= z;
  z %= 1000000007;
  if (p % 2 == 0) return z;
  z *= x;
  z %= 1000000007;
  return z;
}
long long inv(long long x) { return pw(x, 1000000007 - 2); }
long long c(long long x, long long y) {
  if (x < y) swap(x, y);
  if (c1[x][y] != -1) return c1[x][y];
  long long r = fc[y] * fc[x - y];
  r %= 1000000007;
  r = inv(r) * fc[x];
  r %= 1000000007;
  return c1[x][y] = r;
}
int main() {
  memset(c1, -1, sizeof(c1));
  fc[0] = 1;
  for (int i = 1; i < 1010; i++) {
    fc[i] = fc[i - 1] * i;
    fc[i] %= 1000000007;
  }
  for (int i = 0; i < 1010; i++) dp[0][0][i] = 1;
  for (int k = 1; k < 50; k++) {
    for (int i = 1; i < 1010; i++) {
      long long r1 = 0;
      int j = 1;
      for (int pr = 1; pr < 1010; pr++) {
        long long r = 0;
        if (j < pr && i - (j - 1) >= 0) {
          r1 += dp[k - 1][i - (j - 1)][j];
          r1 %= 1000000007;
          j++;
        }
        r += r1;
        r %= 1000000007;
        dp[k][i][pr] = r;
      }
    }
  }
  int t;
  cin >> t;
  int n, k;
  while (t--) {
    scanf("%d%d", &n, &k);
    long long res = 0;
    for (int i = n - (k * (k + 1) / 2); i >= 0; i--) {
      long long r1 = dp[k][n - (i)][1009] * c(i + k, i);
      r1 %= 1000000007;
      r1 *= fc[k];
      r1 %= 1000000007;
      res += r1;
      res %= 1000000007;
    }
    printf("%I64d\n", res);
  }
  return 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;
int dp[50][1010][1010];
long long fc[1010];
long long c1[1010][1010];
long long pw(long long x, long long p) {
  if (p == 0) return 1;
  long long z = pw(x, p / 2);
  z *= z;
  z %= 1000000007;
  if (p % 2 == 0) return z;
  z *= x;
  z %= 1000000007;
  return z;
}
long long inv(long long x) { return pw(x, 1000000007 - 2); }
long long c(long long x, long long y) {
  if (x < y) swap(x, y);
  if (c1[x][y] != -1) return c1[x][y];
  long long r = fc[y] * fc[x - y];
  r %= 1000000007;
  r = inv(r) * fc[x];
  r %= 1000000007;
  return c1[x][y] = r;
}
int main() {
  memset(c1, -1, sizeof(c1));
  fc[0] = 1;
  for (int i = 1; i < 1010; i++) {
    fc[i] = fc[i - 1] * i;
    fc[i] %= 1000000007;
  }
  for (int i = 0; i < 1010; i++) dp[0][0][i] = 1;
  for (int k = 1; k < 50; k++) {
    for (int i = 1; i < 1010; i++) {
      long long r1 = 0;
      int j = 1;
      for (int pr = 1; pr < 1010; pr++) {
        long long r = 0;
        if (j < pr && i - (j - 1) >= 0) {
          r1 += dp[k - 1][i - (j - 1)][j];
          r1 %= 1000000007;
          j++;
        }
        r += r1;
        r %= 1000000007;
        dp[k][i][pr] = r;
      }
    }
  }
  int t;
  cin >> t;
  int n, k;
  while (t--) {
    scanf("%d%d", &n, &k);
    long long res = 0;
    for (int i = n - (k * (k + 1) / 2); i >= 0; i--) {
      long long r1 = dp[k][n - (i)][1009] * c(i + k, i);
      r1 %= 1000000007;
      r1 *= fc[k];
      r1 %= 1000000007;
      res += r1;
      res %= 1000000007;
    }
    printf("%I64d\n", res);
  }
  return 0;
}

```