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>
#pragma GCC optimize(3, "Ofast", "inline")
using namespace std;
const long long inf = 0x3f3f3f3f;
const long long INF = 0x7fffffffffffffff;
const long long mod = 1e9 + 7;
const long long maxn = 1e6 + 5;
long long bit[2005][2005];
int main() {
  for (long long i = 0; i <= 2000; i++) bit[i][0] = 1;
  for (long long i = 1; i <= 2000; i++) {
    for (long long j = 1; j <= 20; j++) {
      bit[i][j] = (bit[i - 1][j] + bit[i - 1][j - 1]) % mod;
    }
  }
  long long n, m;
  cin >> n >> m;
  long long ans = 0;
  for (long long i = 0; i <= n - 1; i++) {
    ans += bit[2 * m - 1 + i][2 * m - 1];
    ans %= mod;
  }
  cout << ans << "\n";
}
