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 MAXINT = 1073741823;
const long long int M = 1000000007;
const int MAXN = 1030;
const long long int MOD = 1000000007;
long long int dp[MAXN][MAXN];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long int a, b, answer;
  cin >> a >> b;
  for (int i = 0; i < MAXN; i++) {
    dp[i][0] = 1;
    dp[i][i] = 1;
    for (int k = 1; k < i; k++) {
      dp[i][k] = (dp[i - 1][k] + dp[i - 1][k - 1]);
      dp[i][k] %= MOD;
    }
  }
  answer = 2 * b - 1 + a;
  cout << dp[answer][a - 1] % MOD;
  return 0;
}
