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;
typedef int64_t ll;
int dpi[10][1001], dpisum[10][1001], dpd[10][1001], dpdsum[10][1002];
int main(int argc, char **argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0), cout.precision(15);
  memset(dpi, 0, sizeof(dpi));
  memset(dpd, 0, sizeof(dpd));
  int n, m;
  cin >> n >> m;
  const int MOD = 1000 * 1000 * 1000 + 7;
  for (int x = 1; x <= n; x++) {
    dpi[0][x] = 1 % MOD;
    dpisum[0][x] = (dpisum[0][x - 1] + dpi[0][x]) % MOD;
  }
  for (int i = 1; i < m; i++) {
    for (int x = 1; x <= n; x++) {
      dpi[i][x] = dpisum[i - 1][x];
      dpisum[i][x] = (dpisum[i][x - 1] + dpi[i][x]) % MOD;
    }
  }
  for (int x = n; x >= 1; x--) {
    dpd[0][x] = 1 % MOD;
    dpdsum[0][x] = (dpdsum[0][x + 1] + dpd[0][x]) % MOD;
  }
  for (int i = 1; i < m; i++) {
    for (int x = n; x >= 1; x--) {
      dpd[i][x] = dpdsum[i - 1][x];
      dpdsum[i][x] = (dpdsum[i][x + 1] + dpd[i][x]) % MOD;
    }
  }
  int ans = 0;
  for (int amax = 1; amax <= n; amax++) {
    for (int bmin = amax; bmin <= n; bmin++) {
      ans =
          (ans + (int)(((ll)dpi[m - 1][amax]) * dpd[m - 1][bmin] % MOD)) % MOD;
    }
  }
  cout << ans << endl;
  return 0;
}
