You are given two integers n and m. Calculate the number of pairs of arrays (a, b) such that:

  * the length of both arrays is equal to m; 
  * each element of each array is an integer between 1 and n (inclusive); 
  * a_i ≤ b_i for any index i from 1 to m; 
  * array a is sorted in non-descending order; 
  * array b is sorted in non-ascending order. 



As the result can be very large, you should print it modulo 10^9+7.

Input

The only line contains two integers n and m (1 ≤ n ≤ 1000, 1 ≤ m ≤ 10).

Output

Print one integer – the number of arrays a and b satisfying the conditions described above modulo 10^9+7.

Examples

Input


2 2


Output


5


Input


10 1


Output


55


Input


723 9


Output


157557417

Note

In the first test there are 5 suitable arrays: 

  * a = [1, 1], b = [2, 2]; 
  * a = [1, 2], b = [2, 2]; 
  * a = [2, 2], b = [2, 2]; 
  * a = [1, 1], b = [2, 1]; 
  * a = [1, 1], b = [1, 1]. 
#include <bits/stdc++.h>
using namespace std;
const int64_t MOD = 1000000007;
int n, m;
int dp[11][1001][1001];
int sums[11][1001][1001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int a = 1; a <= n; a++) {
    for (int b = n; b >= a; b--) {
      dp[1][a][b] = 1;
      sums[1][a][b] =
          (int)(((int64_t)sums[1][a - 1][b] + (int64_t)sums[1][a][b + 1] -
                 (int64_t)sums[1][a - 1][b + 1] + (int64_t)dp[1][a][b]) %
                MOD);
    }
  }
  for (int l = 2; l <= m; l++) {
    for (int a = 1; a <= n; a++) {
      for (int b = n; b >= a; b--) {
        dp[l][a][b] = sums[l - 1][a][b];
        sums[l][a][b] =
            (int)(((int64_t)sums[l][a - 1][b] + (int64_t)sums[l][a][b + 1] -
                   (int64_t)sums[l][a - 1][b + 1] + (int64_t)dp[l][a][b]) %
                  MOD);
      }
    }
  }
  int64_t ans = 0;
  for (int a = 1; a <= n; a++) {
    for (int b = a; b <= n; b++) {
      ans = (ans + (int64_t)dp[m][a][b]) % MOD;
    }
  }
  printf("%" PRId64 "\n", ans);
}
