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 MOD = 1000000000 + 7;
long long binExp(long long x) {
  long long pow = x, ans = 1, power = MOD - 2;
  while (power) {
    if (power & 1) ans = (ans * pow) % MOD;
    pow = (pow * pow) % MOD;
    power >>= 1;
  }
  return ans;
}
int main() {
  long long n, m;
  cin >> n >> m;
  int dp[22][1001];
  for (int i = 0; i < 2 * m; i++) {
    for (int j = 0; j < n; j++) {
      if (i == 0)
        dp[0][j] = j + 1;
      else if (j == 0)
        dp[i][0] = 1;
      else
        dp[i][j] = (dp[i - 1][j] + dp[i][j - 1]) % MOD;
    }
  }
  cout << dp[2 * m - 1][n - 1] << '\n';
  return 0;
}
