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;
long long MOD = 1000000000 + 7;
int N, M;
long long dp[12][1002];
void solve() {
  int i, j, k;
  scanf("%d %d\n", &N, &M);
  for (i = 0; i <= M; i++) dp[i][0] = 1;
  for (i = 0; i <= N; i++) dp[0][i] = 1;
  for (j = 1; j <= M; j++) {
    for (i = 1; i <= N; i++) {
      dp[j][i] = 0;
      for (k = 0; k <= i; k++) {
        dp[j][i] += ((i - k + 1) * dp[j - 1][k]) % MOD;
        dp[j][i] %= MOD;
      }
    }
  }
  printf("%lld\n", dp[M][N - 1]);
}
int main() {
  solve();
  return 0;
}
