#include <bits/stdc++.h>
using namespace std;
const int N = 1000 + 20;
const int MOD = 1000000007;
int k, n;
int fac[N], inv[N];
int C[N][N], p[N][N];
int fpow(int n, int k) {
  int ans = 1, base = n;
  while (k) {
    if (k & 1) ans = (long long)ans * base % MOD;
    base = (long long)base * base % MOD;
    k >>= 1;
  }
  return ans;
}
int add(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
void get_C(int n, int k) {
  C[n][k] = (long long)fac[n] * inv[k] % MOD * inv[n - k] % MOD;
}
void solve() {
  int ans = 0;
  for (int sum = (k) * (k - 1) / 2; sum + (k - 1) <= n; sum++) {
    ans = add(ans, (long long)p[sum + k - 1][k] * C[n - sum + 1][k] % MOD);
  }
  ans = (long long)ans * fac[k] % MOD;
  printf("%d\n", ans);
}
int main() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = (long long)fac[i - 1] * i % MOD;
  inv[0] = 1;
  for (int i = 1; i < N; i++) inv[i] = fpow(fac[i], MOD - 2);
  for (int i = 1; i < N; i++)
    for (int j = 1; j <= i; j++) get_C(i, j);
  p[0][0] = 1;
  for (int i = 1; i < N; i++) {
    for (int j = 1; j <= i; j++) {
      p[i][j] = (p[i - j][j - 1] + p[i - j][j]) % MOD;
    }
  }
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &k);
    solve();
  }
  return 0;
}

### 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;
const int N = 1000 + 20;
const int MOD = 1000000007;
int k, n;
int fac[N], inv[N];
int C[N][N], p[N][N];
int fpow(int n, int k) {
  int ans = 1, base = n;
  while (k) {
    if (k & 1) ans = (long long)ans * base % MOD;
    base = (long long)base * base % MOD;
    k >>= 1;
  }
  return ans;
}
int add(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
void get_C(int n, int k) {
  C[n][k] = (long long)fac[n] * inv[k] % MOD * inv[n - k] % MOD;
}
void solve() {
  int ans = 0;
  for (int sum = (k) * (k - 1) / 2; sum + (k - 1) <= n; sum++) {
    ans = add(ans, (long long)p[sum + k - 1][k] * C[n - sum + 1][k] % MOD);
  }
  ans = (long long)ans * fac[k] % MOD;
  printf("%d\n", ans);
}
int main() {
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = (long long)fac[i - 1] * i % MOD;
  inv[0] = 1;
  for (int i = 1; i < N; i++) inv[i] = fpow(fac[i], MOD - 2);
  for (int i = 1; i < N; i++)
    for (int j = 1; j <= i; j++) get_C(i, j);
  p[0][0] = 1;
  for (int i = 1; i < N; i++) {
    for (int j = 1; j <= i; j++) {
      p[i][j] = (p[i - j][j - 1] + p[i - j][j]) % MOD;
    }
  }
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d%d", &n, &k);
    solve();
  }
  return 0;
}

```