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 NMAX = 1005;
const int MMAX = 15;
const int MOD = 1000000007;
int dp[NMAX][NMAX][MMAX];
int add(int a, int b) {
  a = a + b;
  if (a >= MOD) return a - MOD;
  return a;
}
int main() {
  int n, m, i, j, k, sum;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++)
    for (j = i; j <= n; j++) dp[i][j][1] = 1;
  for (k = 2; k <= m; k++) {
    for (i = 1; i <= n; i++) {
      for (j = n; j >= i; j--) {
        dp[i][j][k] =
            (((dp[i][j][k - 1] + dp[i - 1][j][k]) % MOD + dp[i][j + 1][k]) %
                 MOD +
             MOD - dp[i - 1][j + 1][k]) %
            MOD;
      }
    }
  }
  sum = 0;
  for (i = 1; i <= n; i++)
    for (j = i; j <= n; j++) sum = (sum + dp[i][j][m]) % MOD;
  printf("%d", sum);
  return 0;
}
