#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long maxn = 1e3 + 10, inf = 4e18, mod = 1e9 + 7;
const long double pi = 3.14159265359, eps = 1e-9;
long long fac[maxn + 50], inv[maxn + 50];
long long mdc(long long n) {
  while (n < 0) {
    n += mod;
  }
  return n % mod;
}
long long pwm(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  if (b % 2) {
    return mdc(a * pwm(a, b - 1));
  }
  return pwm(mdc(a * a), b / 2);
}
long long C(long long n, long long r) {
  if (r > n) {
    return 0;
  }
  return mdc(mdc(fac[n] * inv[r]) * inv[n - r]);
}
long long n, k, dp[51][maxn + 50], jj[51][maxn + 50];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fac[0] = inv[0] = 1;
  for (int i = 1; i < maxn + 50; i++) {
    fac[i] = mdc(i * fac[i - 1]);
  }
  for (int i = 1; i < maxn + 50; i++) {
    inv[i] = mdc(inv[i - 1] * pwm(i, mod - 2));
  }
  for (int j = 1; j < maxn + 50; j++) {
    dp[1][j] = 1;
  }
  for (int i = 2; i <= 50; i++)
    for (int j = 2; j < maxn + 50; j++) {
      for (int x = 1; j - x * i >= ceil((long double)i * (i - 1) / 2); x++) {
        dp[i][j] += dp[i - 1][j - x * i];
        dp[i][j] %= mod;
      }
    }
  for (k = 1; k <= 50; k++)
    for (n = 1; n < maxn; n++) {
      for (int sum = 1; sum <= n; sum++) {
        jj[k][n] +=
            (((dp[k][sum] * fac[n - sum + k]) % mod) * inv[n - sum]) % mod;
        jj[k][n] %= mod;
      }
    }
  long long T;
  cin >> T;
  while (T--) {
    cin >> n >> k;
    cout << (k <= 50 ? jj[k][n] : 0) << endl;
  }
  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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long maxn = 1e3 + 10, inf = 4e18, mod = 1e9 + 7;
const long double pi = 3.14159265359, eps = 1e-9;
long long fac[maxn + 50], inv[maxn + 50];
long long mdc(long long n) {
  while (n < 0) {
    n += mod;
  }
  return n % mod;
}
long long pwm(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  if (b % 2) {
    return mdc(a * pwm(a, b - 1));
  }
  return pwm(mdc(a * a), b / 2);
}
long long C(long long n, long long r) {
  if (r > n) {
    return 0;
  }
  return mdc(mdc(fac[n] * inv[r]) * inv[n - r]);
}
long long n, k, dp[51][maxn + 50], jj[51][maxn + 50];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fac[0] = inv[0] = 1;
  for (int i = 1; i < maxn + 50; i++) {
    fac[i] = mdc(i * fac[i - 1]);
  }
  for (int i = 1; i < maxn + 50; i++) {
    inv[i] = mdc(inv[i - 1] * pwm(i, mod - 2));
  }
  for (int j = 1; j < maxn + 50; j++) {
    dp[1][j] = 1;
  }
  for (int i = 2; i <= 50; i++)
    for (int j = 2; j < maxn + 50; j++) {
      for (int x = 1; j - x * i >= ceil((long double)i * (i - 1) / 2); x++) {
        dp[i][j] += dp[i - 1][j - x * i];
        dp[i][j] %= mod;
      }
    }
  for (k = 1; k <= 50; k++)
    for (n = 1; n < maxn; n++) {
      for (int sum = 1; sum <= n; sum++) {
        jj[k][n] +=
            (((dp[k][sum] * fac[n - sum + k]) % mod) * inv[n - sum]) % mod;
        jj[k][n] %= mod;
      }
    }
  long long T;
  cin >> T;
  while (T--) {
    cin >> n >> k;
    cout << (k <= 50 ? jj[k][n] : 0) << endl;
  }
  return 0;
}

```