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;
template <typename t>
void read(t &x) {
  char ch = getchar();
  x = 0;
  t f = 1;
  while (ch < '0' || ch > '9') f = (ch == '-' ? -1 : f), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
long long dp[1000][100];
int main() {
  int n, m;
  read(n), read(m);
  m *= 2;
  ;
  for (int i = 1; i <= n; i++) dp[1][i] = 1;
  for (int i = 2; i <= m; i++)
    for (int j = n; j > 0; j--)
      dp[i][j] = (dp[i][j + 1] + dp[i - 1][j]) % 1000000007;
  long long ans = 0;
  for (int i = 1; i <= n; i++) ans = (ans + dp[m][i]) % 1000000007;
  cout << ans << endl;
  return 0;
}
