#include <bits/stdc++.h>
const int N = 1005, mod = 1e9 + 7;
int n, k, t;
long long f[N][N], fac[N] = {1}, inv[N], ans[N][N];
inline int qpow(long long base, int exp) {
  long long res = 1;
  while (exp) {
    if (exp & 1) res = res * base % mod;
    base = base * base % mod;
    exp >>= 1;
  }
  return res;
}
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline long long A(int n, int m) {
  if (n < m) return 0;
  return fac[n] * inv[n - m] % mod;
}
int main() {
  scanf("%d", &t);
  for (int i = 1; i < N; ++i) fac[i] = fac[i - 1] * i % mod;
  inv[N - 1] = qpow(fac[N - 1], mod - 2), inv[0] = 1;
  for (int i = N - 2; i; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
  f[0][0] = 1;
  for (int i = 1; i <= 45; ++i) {
    for (int j = (i + 1) * i >> 1; j < N; ++j)
      f[i][j] = add(f[i][j - i], f[i - 1][j - i]);
  }
  for (int i = 1; i < N; ++i) {
    for (int j = 1; j <= 45; ++j) {
      for (int k = j * (j + 1) >> 1; k <= i; ++k)
        ans[i][j] += 1ll * f[j][k] * A(i - k + j, j) % mod, ans[i][j] %= mod;
    }
  }
  while (t--) {
    scanf("%d%d", &n, &k);
    printf("%lld\n", ans[n][k]);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
const int N = 1005, mod = 1e9 + 7;
int n, k, t;
long long f[N][N], fac[N] = {1}, inv[N], ans[N][N];
inline int qpow(long long base, int exp) {
  long long res = 1;
  while (exp) {
    if (exp & 1) res = res * base % mod;
    base = base * base % mod;
    exp >>= 1;
  }
  return res;
}
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline long long A(int n, int m) {
  if (n < m) return 0;
  return fac[n] * inv[n - m] % mod;
}
int main() {
  scanf("%d", &t);
  for (int i = 1; i < N; ++i) fac[i] = fac[i - 1] * i % mod;
  inv[N - 1] = qpow(fac[N - 1], mod - 2), inv[0] = 1;
  for (int i = N - 2; i; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
  f[0][0] = 1;
  for (int i = 1; i <= 45; ++i) {
    for (int j = (i + 1) * i >> 1; j < N; ++j)
      f[i][j] = add(f[i][j - i], f[i - 1][j - i]);
  }
  for (int i = 1; i < N; ++i) {
    for (int j = 1; j <= 45; ++j) {
      for (int k = j * (j + 1) >> 1; k <= i; ++k)
        ans[i][j] += 1ll * f[j][k] * A(i - k + j, j) % mod, ans[i][j] %= mod;
    }
  }
  while (t--) {
    scanf("%d%d", &n, &k);
    printf("%lld\n", ans[n][k]);
  }
  return 0;
}

```