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 mod = 1e9 + 7;
long long Bpow(long long a, long long s) {
  if (s == 0) {
    return 1;
  }
  long long u = Bpow(a, s / 2);
  if (s & 1)
    return u * u % mod * a % mod;
  else
    return u * u % mod;
}
int32_t main() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> dpa(m, vector<long long>(n + 1));
  for (long long i = 1; i < n + 1; i++) {
    dpa[0][i] = dpa[0][i - 1] + 1;
  }
  for (long long i = 1; i < m; i++) {
    for (long long j = 1; j < n + 1; j++) {
      dpa[i][j] = (dpa[i - 1][j] + dpa[i][j - 1]) % mod;
    }
  }
  vector<vector<long long>> dpb(m, vector<long long>(n + 2));
  for (long long i = n; i > 0; i--) {
    dpb[0][i] = dpb[0][i + 1] + 1;
  }
  for (long long i = 1; i < m; i++) {
    for (long long j = n; j > 0; j--) {
      dpb[i][j] = (dpb[i - 1][j] + dpb[i][j + 1]) % mod;
    }
  }
  long long ans = 0;
  for (long long i = 1; i < n + 1; i++) {
    ans +=
        ((dpa[m - 1][i] - dpa[m - 1][i - 1]) + mod) % mod * dpb[m - 1][i] % mod;
    ans %= mod;
  }
  cout << ans;
}
