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 N = 1005;
const int M = 11;
const int MOD = 1e9 + 7;
int n, m;
int dp[M][N][N];
int solve(int idx, int a, int b) {
  if (idx >= m) return 1;
  int& r = dp[idx][a][b];
  if (r != -1) return r;
  r = solve(idx + 1, a, b);
  if (a < b) {
    r = (r + solve(idx, a + 1, b)) % MOD;
    r = (r + solve(idx, a, b - 1)) % MOD;
  }
  if (a + 1 < b) r = ((r - solve(idx, a + 1, b - 1)) % MOD + MOD) % MOD;
  return r;
}
int main() {
  cin >> n >> m;
  memset(dp, -1, sizeof dp);
  cout << solve(0, 1, n) << '\n';
  return 0;
}
