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 = 1000000007;
int n, m, dp[12][1005][1005];
int f(int i, int a, int b) {
  if (a < 1 || b > n) return 0;
  if (i == m + 1) return 1;
  if (dp[i][a][b] != -1) return dp[i][a][b];
  int ans = (f(i, a + 1, b) + f(i, a, b - 1)) % MOD;
  ans = (ans - f(i, a + 1, b - 1) + MOD) % MOD;
  if (a <= b) ans = (ans + f(i + 1, a, b)) % MOD;
  return dp[i][a][b] = ans;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++)
    for (int j = 1; j <= n; j++)
      for (int k = 1; k <= n; k++) dp[i][j][k] = -1;
  printf("%d\n", f(1, 1, n));
  return 0;
}
