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 long long mod = (long long)1e9 + 7;
long long n, m, t, ans;
long long dp[1010][1010];
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  cin >> n >> m;
  dp[0][0] = 1;
  for (long long i = 1; i <= m; i++) {
    for (long long j = 1; j <= n; j++) {
      for (long long k = 0; k <= j; k++) {
        dp[i][j] += dp[i - 1][k];
        dp[i][j] %= mod;
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j++) {
      ans += (dp[m][n - j + 1] * dp[m][i]) % mod;
      ans %= mod;
    }
  }
  cout << ans;
}
