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 MAX_N = 1e3 + 5;
const long long mod = 1e9 + 7;
int n, m;
long long dp[15][MAX_N], sum[15][MAX_N], sum2[MAX_N];
int main() {
  memset(sum, 0, sizeof(sum)), memset(dp, 0, sizeof(dp));
  scanf("%d", &n), scanf("%d", &m);
  for (register int i = 1; i <= n; i++)
    dp[1][i] = 1ll, sum[1][i] = dp[1][i] + sum[1][i - 1];
  for (register int i = 2; i <= m; i++)
    for (register int j = 1; j <= n; j++) {
      dp[i][j] = sum[i - 1][j];
      sum[i][j] = (sum[i][j - 1] + dp[i][j]) % mod;
    }
  for (register int i = 1; i <= n; i++)
    sum2[i] = (sum2[i - 1] + dp[m][i]) % mod;
  long long ans = 0;
  for (register int i = 1; i <= n; i++)
    ans = (ans + (dp[m][i] * sum2[n - i + 1]) % mod) % mod;
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}
