#include <bits/stdc++.h>
using namespace std;
const int N = 1010, mod = 1e9 + 7;
int t, n, k;
long long fac[N], inv[N], f[60][N], ans[N][60];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline long long mul(long long a, long long b) {
  return a * (long long)b % mod;
}
inline long long add(long long a, long long b) {
  return a + b > mod ? a + b - mod : a + b;
}
inline long long fpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline void init() {
  fac[0] = 1;
  for (int i = 1; i <= 1000; i++) fac[i] = mul(fac[i - 1], i);
  inv[1000] = fpow(fac[1000], mod - 2);
  for (int i = 1000; i > 0; i--) inv[i - 1] = mul(inv[i], i);
}
inline long long C(long long n, long long m) {
  if (n < m) return 0;
  return mul(fac[n], mul(inv[n - m], inv[m]));
}
int main() {
  init();
  f[0][0] = 1;
  for (int i = 1; i <= 50; i++)
    for (int j = i * (i + 1) / 2; j <= 1000; j++)
      f[i][j] = add(f[i][j - i], f[i - 1][j - i]);
  for (int i = 1; i <= 1000; i++)
    for (int k = 1; k <= 50; k++) {
      for (int j = k * (k + 1) / 2; j <= i; j++) {
        ans[i][k] = add(ans[i][k], mul(C(i - j + k, k), f[k][j]));
      }
      ans[i][k] = mul(ans[i][k], fac[k]);
    }
  t = read();
  while (t--) {
    n = read();
    k = read();
    if (k > 50)
      puts("0");
    else
      printf("%lld\n", ans[n][k]);
  }
  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 = 1010, mod = 1e9 + 7;
int t, n, k;
long long fac[N], inv[N], f[60][N], ans[N][60];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
inline long long mul(long long a, long long b) {
  return a * (long long)b % mod;
}
inline long long add(long long a, long long b) {
  return a + b > mod ? a + b - mod : a + b;
}
inline long long fpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline void init() {
  fac[0] = 1;
  for (int i = 1; i <= 1000; i++) fac[i] = mul(fac[i - 1], i);
  inv[1000] = fpow(fac[1000], mod - 2);
  for (int i = 1000; i > 0; i--) inv[i - 1] = mul(inv[i], i);
}
inline long long C(long long n, long long m) {
  if (n < m) return 0;
  return mul(fac[n], mul(inv[n - m], inv[m]));
}
int main() {
  init();
  f[0][0] = 1;
  for (int i = 1; i <= 50; i++)
    for (int j = i * (i + 1) / 2; j <= 1000; j++)
      f[i][j] = add(f[i][j - i], f[i - 1][j - i]);
  for (int i = 1; i <= 1000; i++)
    for (int k = 1; k <= 50; k++) {
      for (int j = k * (k + 1) / 2; j <= i; j++) {
        ans[i][k] = add(ans[i][k], mul(C(i - j + k, k), f[k][j]));
      }
      ans[i][k] = mul(ans[i][k], fac[k]);
    }
  t = read();
  while (t--) {
    n = read();
    k = read();
    if (k > 50)
      puts("0");
    else
      printf("%lld\n", ans[n][k]);
  }
  return 0;
}

```