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 MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  int dp[n][n], new_dp[n][n], suf[n][n], val[n][n];
  memset(dp, 0, sizeof(dp));
  memset(new_dp, 0, sizeof(new_dp));
  memset(suf, 0, sizeof(suf));
  memset(val, 0, sizeof(val));
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      dp[i][j] += 1;
    }
  }
  for (int j = 0; j < n; j++) {
    for (int k = n - 1; k >= j; k--) {
      suf[j][k] = (k + 1 < n ? suf[j][k + 1] : 0) + dp[j][k];
      suf[j][k] %= MOD;
    }
    for (int k = n - 1; k >= j; k--) {
      val[j][k] = (j - 1 >= 0 ? val[j - 1][k] : 0) + suf[j][k];
      val[j][k] %= MOD;
    }
  }
  for (int i = 2; i <= m; i++) {
    for (int jj = 0; jj < n; jj++) {
      for (int kk = jj; kk < n; kk++) {
        new_dp[jj][kk] = val[jj][kk];
      }
    }
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        dp[j][k] = new_dp[j][k];
        new_dp[j][k] = 0;
      }
    }
    for (int j = 0; j < n; j++) {
      for (int k = n - 1; k >= j; k--) {
        suf[j][k] = (k + 1 < n ? suf[j][k + 1] : 0) + dp[j][k];
        suf[j][k] %= MOD;
      }
      for (int k = n - 1; k >= j; k--) {
        val[j][k] = (j - 1 >= 0 ? val[j - 1][k] : 0) + suf[j][k];
        val[j][k] %= MOD;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i; j < n; j++) {
      ans += dp[i][j];
      ans %= MOD;
    }
  }
  cout << ans;
}
