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 = 1000000000000005;
const long long maxn = 5e5 + 5;
const int mod = 1e9 + 7;
long long n, m, p;
long long num[maxn];
long long dp[30][maxn];
long long mx = 0;
int main() {
  scanf("%lld%lld", &n, &m);
  dp[0][1] = 1;
  for (int i = 1; i <= 2 * m; i++) {
    for (int k = 1; k <= n; k++) {
      for (int j = 1; j <= k; j++) dp[i][k] = (dp[i][k] + dp[i - 1][j]) % mod;
    }
  }
  long long res = 0;
  for (int i = 1; i <= n; i++) res = (res + dp[2 * m][i]) % mod;
  printf("%lld\n", res);
  return 0;
}
