#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 2e3 + 10;
const long long mod = 1e9 + 7;
inline long long bpow(long long t, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * t) % mod;
    t = (t * t) % mod, n >>= 1;
  }
  return ans;
}
long long fact[MAX_N], ifact[MAX_N];
inline long long c(long long n, long long k) {
  if (k > n || k < 0) return 0;
  return (fact[n] * ifact[k] % mod * ifact[n - k]) % mod;
}
inline void get_fact() {
  fact[0] = 1;
  ifact[0] = 1;
  for (long long i = 1; i < MAX_N; i++)
    fact[i] = (fact[i - 1] * i) % mod, ifact[i] = bpow(fact[i], mod - 2);
}
int dp[MAX_N][MAX_N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(0);
  get_fact();
  for (int n = 0; n <= 1000; n++) dp[0][n] = 1;
  for (int k = 1; k < 45; k++) {
    for (int n = k; n <= 1000; n++) {
      int sum = dp[k][n - k];
      if (k == 1) sum += 1;
      for (int n2 = 1; n2 <= n - k; n2++) sum += dp[k - 1][n2], sum %= mod;
      dp[k][n] = sum;
      dp[k][n] %= mod;
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int k, n;
    cin >> n >> k;
    if (k >= 45)
      cout << 0;
    else
      cout << fact[k] * dp[k][n] % mod;
    cout << endl;
  }
}

### Prompt

Your challenge is to write 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 MAX_N = 2e3 + 10;
const long long mod = 1e9 + 7;
inline long long bpow(long long t, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * t) % mod;
    t = (t * t) % mod, n >>= 1;
  }
  return ans;
}
long long fact[MAX_N], ifact[MAX_N];
inline long long c(long long n, long long k) {
  if (k > n || k < 0) return 0;
  return (fact[n] * ifact[k] % mod * ifact[n - k]) % mod;
}
inline void get_fact() {
  fact[0] = 1;
  ifact[0] = 1;
  for (long long i = 1; i < MAX_N; i++)
    fact[i] = (fact[i - 1] * i) % mod, ifact[i] = bpow(fact[i], mod - 2);
}
int dp[MAX_N][MAX_N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(0);
  get_fact();
  for (int n = 0; n <= 1000; n++) dp[0][n] = 1;
  for (int k = 1; k < 45; k++) {
    for (int n = k; n <= 1000; n++) {
      int sum = dp[k][n - k];
      if (k == 1) sum += 1;
      for (int n2 = 1; n2 <= n - k; n2++) sum += dp[k - 1][n2], sum %= mod;
      dp[k][n] = sum;
      dp[k][n] %= mod;
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int k, n;
    cin >> n >> k;
    if (k >= 45)
      cout << 0;
    else
      cout << fact[k] * dp[k][n] % mod;
    cout << endl;
  }
}

```