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 N = 1005;
long long n, m;
long long f[N][N], g[N][N], ans;
;
inline long long read() {
  long long x = 0, y = 1;
  char ch = '\000';
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') y = -1;
  for (; isdigit(ch); ch = getchar()) x = (x << 3) + (x << 1) + (ch ^ '0');
  return x * y;
}
void work() {
  for (long long i = 1; i <= n; i++) f[1][i] = g[1][i] = 1;
  for (register long long i = 2; i <= m; i++)
    for (register long long j = 1; j <= n; j++)
      for (register long long k = 1; k <= j; k++)
        f[i][j] = (f[i][j] + f[i - 1][k]) % 1000000007;
  for (register long long i = 2; i <= m; i++)
    for (register long long j = 1; j <= n; j++)
      for (register long long k = j; k <= n; k++)
        g[i][j] = (g[i][j] + g[i - 1][k]) % 1000000007;
}
signed main() {
  n = read();
  m = read();
  work();
  for (register long long i = 1; i <= n; i++)
    for (register long long j = 1; j <= i; j++)
      ans = (ans + f[m][j] * g[m][i]) % 1000000007;
  printf("%lld\n", ans);
  return 0;
}
