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 long long INF = 2e18;
const long long N = 1e5 + 5;
const long long MOD = 1e9 + 7;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
  return x * f;
}
long long dp[1005][1005], n, m, sum[1005][1005];
int main() {
  scanf("%lld%lld", &n, &m);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      sum[i][j] = 1;
    }
  }
  for (int k = 2; k <= m; k++) {
    for (int i = n; i >= 1; i--) {
      for (int j = 1; j <= i; j++) {
        sum[i][j] = (sum[i][j] + sum[i + 1][j] + sum[i][j - 1] -
                     sum[i + 1][j - 1] + MOD) %
                    MOD;
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      ans += sum[i][j];
      ans %= MOD;
    }
  }
  cout << ans << endl;
}
