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;
long long int exp(long long int a, long long int b) {
  long long int res = 1LL;
  while (b > 0) {
    if (b & 1) {
      res = res * a;
    }
    a = (a * a);
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, m, res = 0LL;
  cin >> n >> m;
  long long int dp[m][n + 1][n + 1];
  for (long long int la = 1; la < n + 1; ++la) {
    for (long long int lb = la; lb < n + 1; ++lb) {
      dp[0][la][lb] = 1LL;
    }
  }
  for (long long int idx = 1; idx < m; ++idx) {
    for (long long int la = 1; la < n + 1; ++la) {
      for (long long int lb = la; lb < n + 1; ++lb) {
        dp[idx][la][lb] = dp[idx - 1][la][lb];
      }
    }
    for (long long int a = 2; a < n + 1; ++a) {
      for (long long int b = a; b < n + 1; ++b) {
        dp[idx][a][b] = (dp[idx][a][b] + dp[idx][a - 1][b]) % 1000000007;
      }
    }
    for (long long int b = n - 1; b >= 1; --b) {
      for (long long int a = 1; a < b + 1; ++a) {
        dp[idx][a][b] = (dp[idx][a][b] + dp[idx][a][b + 1]) % 1000000007;
      }
    }
  }
  for (long long int la = 1; la < n + 1; ++la) {
    for (long long int lb = la; lb < n + 1; ++lb) {
      res = (res + dp[m - 1][la][lb]) % 1000000007;
    }
  }
  cout << res << endl;
  return 0;
}
