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>
#pragma GCC optimize("O3")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long mod = 1e9 + 7;
long long dp[1000 + 1][22];
int32_t main() {
  ios_base::sync_with_stdio(false);
  long long n, m;
  cin >> n >> m;
  m *= 2;
  dp[1][0] = 1;
  long long ans = 0;
  for (long long i = 1; i <= m; i++) {
    long long sum = 0;
    for (long long x = 1; x <= n; x++) {
      sum += dp[x][i - 1];
      if (sum >= mod) sum -= mod;
      dp[x][i] = sum;
    }
  }
  for (long long x = 1; x <= n; x++) {
    ans += dp[x][m];
    if (ans >= mod) ans -= mod;
  }
  cout << ans << "\n";
}
