#include <bits/stdc++.h>
using namespace std;
long long mem[1003][1003], fact[2 * 1003], inv[2 * 1003];
long long dp(long long r1, long long r2) {
  if (!r2) return (r1 == 0);
  if (r1 < r2) return 0;
  if (mem[r1][r2] != -1) return mem[r1][r2];
  return mem[r1][r2] = (dp(r1 - r2, r2) + dp(r1 - r2, r2 - 1)) % 1000000007ll;
}
long long bigmod(long long b, long long p) {
  if (!p) return 1ll;
  long long ret = bigmod(b, p / 2);
  ret = (ret * ret) % 1000000007ll;
  if (p & 1ll) ret = (ret * b) % 1000000007ll;
  return ret;
}
long long ncr(int n, int r) {
  return (fact[n] * ((inv[r] * inv[n - r]) % 1000000007ll)) % 1000000007ll;
}
int main() {
  fact[0] = 1ll;
  for (int i = 1; i <= 2000; i++) fact[i] = (fact[i - 1] * i) % 1000000007ll;
  inv[2000] = bigmod(fact[2000], 1000000007ll - 2);
  for (int i = 2000 - 1; i >= 0; i--)
    inv[i] = (inv[i + 1] * (i + 1)) % 1000000007ll;
  int ts;
  scanf("%d", &ts);
  memset(mem, -1, sizeof mem);
  while (ts--) {
    int n, k;
    scanf("%d %d", &n, &k);
    long long res = 0;
    for (int i = k; i <= n; i++) {
      long long t = (dp(i, k) * fact[k]) % 1000000007ll;
      long long rem = n - i;
      t = (t * ncr(rem + k, k)) % 1000000007ll;
      res += t;
      res %= 1000000007ll;
    }
    printf("%lld\n", res);
  }
  return 0;
}

### Prompt

Develop a solution in CPP 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;
long long mem[1003][1003], fact[2 * 1003], inv[2 * 1003];
long long dp(long long r1, long long r2) {
  if (!r2) return (r1 == 0);
  if (r1 < r2) return 0;
  if (mem[r1][r2] != -1) return mem[r1][r2];
  return mem[r1][r2] = (dp(r1 - r2, r2) + dp(r1 - r2, r2 - 1)) % 1000000007ll;
}
long long bigmod(long long b, long long p) {
  if (!p) return 1ll;
  long long ret = bigmod(b, p / 2);
  ret = (ret * ret) % 1000000007ll;
  if (p & 1ll) ret = (ret * b) % 1000000007ll;
  return ret;
}
long long ncr(int n, int r) {
  return (fact[n] * ((inv[r] * inv[n - r]) % 1000000007ll)) % 1000000007ll;
}
int main() {
  fact[0] = 1ll;
  for (int i = 1; i <= 2000; i++) fact[i] = (fact[i - 1] * i) % 1000000007ll;
  inv[2000] = bigmod(fact[2000], 1000000007ll - 2);
  for (int i = 2000 - 1; i >= 0; i--)
    inv[i] = (inv[i + 1] * (i + 1)) % 1000000007ll;
  int ts;
  scanf("%d", &ts);
  memset(mem, -1, sizeof mem);
  while (ts--) {
    int n, k;
    scanf("%d %d", &n, &k);
    long long res = 0;
    for (int i = k; i <= n; i++) {
      long long t = (dp(i, k) * fact[k]) % 1000000007ll;
      long long rem = n - i;
      t = (t * ncr(rem + k, k)) % 1000000007ll;
      res += t;
      res %= 1000000007ll;
    }
    printf("%lld\n", res);
  }
  return 0;
}

```