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 int mod = 1e9 + 7;
const int maxn = 1005;
int f1[maxn][15], f2[maxn][15];
int pre[maxn], suf[maxn];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  f1[0][0] = 1;
  for (int len = 1; len <= m; len++) {
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= i; j++) {
        f1[i][len] = ((long long)f1[i][len] + f1[j][len - 1]) % mod;
      }
    }
  }
  f2[n + 1][0] = 1;
  for (int len = 1; len <= m; len++) {
    for (int i = n; i >= 1; i--) {
      for (int j = n + 1; j >= i; j--) {
        f2[i][len] = ((long long)f2[i][len] + f2[j][len - 1]) % mod;
      }
    }
  }
  for (int i = n; i >= 1; i--)
    suf[i] = ((long long)suf[i + 1] + f2[i][m]) % mod;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = (ans + (long long)f1[i][m] * suf[i]) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
