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 n, m, dp[1005][1005][10];
long long fn(long long l, long long r, long long pos) {
  if (pos == m) return 1;
  if (l > r) return 0;
  long long &z = dp[l][r][pos];
  if (z != -1) return z;
  z = 0;
  z += fn(l, r, pos + 1);
  z %= 1000000007;
  z += fn(l + 1, r, pos);
  z %= 1000000007;
  z += fn(l, r - 1, pos);
  z %= 1000000007;
  z -= fn(l + 1, r - 1, pos);
  z += 1000000007;
  z %= 1000000007;
  return z;
}
int main() {
  long long int i, j, k, z = 0, c = 0, t, x, y, l, r, mid;
  scanf("%lld", &n);
  scanf("%lld", &m);
  memset(dp, -1, sizeof(dp));
  z = fn(1, n, 0);
  cout << z << endl;
}
