#include <bits/stdc++.h>
using namespace std;
const int N = 1200;
int dp[52][1002][1002];
long long fac[N], Inv[N];
const int Mod = 1e9 + 7;
void Add(int &a, long long b) { a = (1ll * a + b % Mod) % Mod; }
long long Pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % Mod;
    a = a * a % Mod;
    b >>= 1;
  }
  return ret;
}
void Init() {
  fac[0] = Inv[0] = 1;
  for (int i = 1; i <= N - 20; i++) fac[i] = fac[i - 1] * i % Mod;
  for (int i = 1; i <= N - 20; i++) Inv[i] = Pow(i, Mod - 2);
  for (int i = 1; i <= N - 20; i++) Inv[i] = Inv[i - 1] * Inv[i] % Mod;
}
long long C(long long a, long long b) {
  if (b > a) return 0;
  return fac[a] * Inv[b] % Mod * Inv[a - b] % Mod;
}
int main() {
  Init();
  dp[0][0][0] = 1;
  for (int i = 0; i <= 50; i++) {
    for (int j = 0; j <= 1000; j++) {
      for (int k = 0; k < 1000; k++) {
        Add(dp[i][j][k + 1], dp[i][j][k]);
      }
    }
    for (int j = 0; j <= 1000; j++) {
      for (int k = 0; k < 1000; k++) {
        if (j + k <= 1000) Add(dp[i + 1][j + k][k + 1], dp[i][j][k]);
      }
    }
  }
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k > 50) {
      printf("0\n");
      continue;
    }
    int ans = 0;
    for (int i = 0; i <= n; i++)
      Add(ans, 1ll * dp[k][i][n] * fac[k] % Mod * C(n - i, k) % Mod);
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Generate 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;
const int N = 1200;
int dp[52][1002][1002];
long long fac[N], Inv[N];
const int Mod = 1e9 + 7;
void Add(int &a, long long b) { a = (1ll * a + b % Mod) % Mod; }
long long Pow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % Mod;
    a = a * a % Mod;
    b >>= 1;
  }
  return ret;
}
void Init() {
  fac[0] = Inv[0] = 1;
  for (int i = 1; i <= N - 20; i++) fac[i] = fac[i - 1] * i % Mod;
  for (int i = 1; i <= N - 20; i++) Inv[i] = Pow(i, Mod - 2);
  for (int i = 1; i <= N - 20; i++) Inv[i] = Inv[i - 1] * Inv[i] % Mod;
}
long long C(long long a, long long b) {
  if (b > a) return 0;
  return fac[a] * Inv[b] % Mod * Inv[a - b] % Mod;
}
int main() {
  Init();
  dp[0][0][0] = 1;
  for (int i = 0; i <= 50; i++) {
    for (int j = 0; j <= 1000; j++) {
      for (int k = 0; k < 1000; k++) {
        Add(dp[i][j][k + 1], dp[i][j][k]);
      }
    }
    for (int j = 0; j <= 1000; j++) {
      for (int k = 0; k < 1000; k++) {
        if (j + k <= 1000) Add(dp[i + 1][j + k][k + 1], dp[i][j][k]);
      }
    }
  }
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    scanf("%d %d", &n, &k);
    if (k > 50) {
      printf("0\n");
      continue;
    }
    int ans = 0;
    for (int i = 0; i <= n; i++)
      Add(ans, 1ll * dp[k][i][n] * fac[k] % Mod * C(n - i, k) % Mod);
    printf("%d\n", ans);
  }
  return 0;
}

```