#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const int MAXN = 1e3 + 10;
inline long long in() {
  long long x = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * flag;
}
int T;
int n, m;
long long ans;
long long fac[MAXN];
long long c[MAXN << 1][MAXN << 1];
long long dp[52][MAXN];
void prework() {
  fac[0] = 1;
  for (int i = 1; i <= 1000; i++) fac[i] = fac[i - 1] * 1LL * i % MOD;
  c[0][0] = 1;
  for (int i = 1; i <= 2000; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      (c[i][j] = c[i - 1][j - 1] + c[i - 1][j]) %= MOD;
  }
  dp[0][0] = 1;
  for (int i = 0; i <= 50; i++) {
    for (int j = i * (i + 1) / 2; j <= 1000; j++) {
      if (j + i <= 1000 && i != 0) (dp[i][j + i] += dp[i][j]) %= MOD;
      if (j + i + 1 <= 1000) (dp[i + 1][j + i + 1] += dp[i][j]) %= MOD;
    }
  }
}
int main() {
  T = in();
  prework();
  while (T--) {
    ans = 0;
    n = in(), m = in();
    for (int i = m * (m + 1) / 2; i <= n; i++) {
      if (dp[m][i] == 0) continue;
      int res = n - i;
      (ans += c[res + m][m] * fac[m] % MOD * dp[m][i] % MOD) %= MOD;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 = 1e9 + 7;
const int MAXN = 1e3 + 10;
inline long long in() {
  long long x = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * flag;
}
int T;
int n, m;
long long ans;
long long fac[MAXN];
long long c[MAXN << 1][MAXN << 1];
long long dp[52][MAXN];
void prework() {
  fac[0] = 1;
  for (int i = 1; i <= 1000; i++) fac[i] = fac[i - 1] * 1LL * i % MOD;
  c[0][0] = 1;
  for (int i = 1; i <= 2000; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      (c[i][j] = c[i - 1][j - 1] + c[i - 1][j]) %= MOD;
  }
  dp[0][0] = 1;
  for (int i = 0; i <= 50; i++) {
    for (int j = i * (i + 1) / 2; j <= 1000; j++) {
      if (j + i <= 1000 && i != 0) (dp[i][j + i] += dp[i][j]) %= MOD;
      if (j + i + 1 <= 1000) (dp[i + 1][j + i + 1] += dp[i][j]) %= MOD;
    }
  }
}
int main() {
  T = in();
  prework();
  while (T--) {
    ans = 0;
    n = in(), m = in();
    for (int i = m * (m + 1) / 2; i <= n; i++) {
      if (dp[m][i] == 0) continue;
      int res = n - i;
      (ans += c[res + m][m] * fac[m] % MOD * dp[m][i] % MOD) %= MOD;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```