#include <bits/stdc++.h>
using namespace std;
int dp[2510][2510];
long long sil[2510], od[2510];
const int P = 1000 * 1000 * 1000 + 7;
long long pot(long long a, int wyk) {
  long long wyn = 1;
  while (wyk) {
    if (wyk & 1) {
      wyn *= a;
      wyn %= P;
    }
    wyk /= 2;
    a *= a;
    a %= P;
  }
  return wyn;
}
long long po(int a, int b) { return sil[a] * od[b] % P * od[a - b] % P; }
int test(int n, int k) {
  if (k * (k + 1) / 2 > n) return 0;
  int wyn = 0;
  for (int i = 0; i < n; i++) {
    wyn += po(k + i, i) * dp[n - i][k] % P;
    if (wyn >= P) wyn -= P;
  }
  return wyn * sil[k] % P;
}
int wyn[2510][2510];
int main() {
  sil[0] = 1;
  od[0] = 1;
  for (int i = 1; i < 2510; i++) {
    sil[i] = sil[i - 1] * i % P;
    od[i] = pot(sil[i], P - 2);
  }
  dp[0][0] = 1;
  for (int i = 1; i < 2510; i++)
    for (int j = 1; j < i + 1; j++) {
      dp[i][j] = dp[i - j][j] + dp[i - j][j - 1];
      if (dp[i][j] >= P) dp[i][j] -= P;
    }
  for (int i = 0; i < 1001; i++)
    for (int j = 0; j < 1001; j++) wyn[i][j] = test(i, j);
  int _;
  scanf("%d", &_);
  while (_--) {
    int i, j;
    scanf("%d", &i);
    scanf("%d", &j);
    printf("%d\n", wyn[i][j]);
  }
}

### Prompt

Please provide a cpp coded solution 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;
int dp[2510][2510];
long long sil[2510], od[2510];
const int P = 1000 * 1000 * 1000 + 7;
long long pot(long long a, int wyk) {
  long long wyn = 1;
  while (wyk) {
    if (wyk & 1) {
      wyn *= a;
      wyn %= P;
    }
    wyk /= 2;
    a *= a;
    a %= P;
  }
  return wyn;
}
long long po(int a, int b) { return sil[a] * od[b] % P * od[a - b] % P; }
int test(int n, int k) {
  if (k * (k + 1) / 2 > n) return 0;
  int wyn = 0;
  for (int i = 0; i < n; i++) {
    wyn += po(k + i, i) * dp[n - i][k] % P;
    if (wyn >= P) wyn -= P;
  }
  return wyn * sil[k] % P;
}
int wyn[2510][2510];
int main() {
  sil[0] = 1;
  od[0] = 1;
  for (int i = 1; i < 2510; i++) {
    sil[i] = sil[i - 1] * i % P;
    od[i] = pot(sil[i], P - 2);
  }
  dp[0][0] = 1;
  for (int i = 1; i < 2510; i++)
    for (int j = 1; j < i + 1; j++) {
      dp[i][j] = dp[i - j][j] + dp[i - j][j - 1];
      if (dp[i][j] >= P) dp[i][j] -= P;
    }
  for (int i = 0; i < 1001; i++)
    for (int j = 0; j < 1001; j++) wyn[i][j] = test(i, j);
  int _;
  scanf("%d", &_);
  while (_--) {
    int i, j;
    scanf("%d", &i);
    scanf("%d", &j);
    printf("%d\n", wyn[i][j]);
  }
}

```