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 mod = 1000000007;
int gcdExtended(int a, int b, int* x, int* y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long modInverse(long long a, long long m) {
  int x, y;
  int g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return 0;
  else {
    long long res = (x % m + m) % m;
    return res;
  }
}
long long nCr(int n, int r) {
  if (r > n) {
    return 0;
  }
  if (r > n - r) {
    r = n - r;
  }
  long long ans = 1;
  for (int i = 1; i <= r; i++) {
    ans *= (n - i + 1);
    ans %= mod;
    ans *= modInverse(i, mod);
    ans %= mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  long long ans = 0;
  ans = nCr(n + 2 * m - 1, 2 * m);
  ans %= mod;
  cout << ans;
}
