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 MD = 1e9 + 7;
long long fpow(long long base, long long power, long long md) {
  long long result = 1;
  while (power > 0) {
    if (power & 1) result = (result * base) % md;
    base = (base * base) % md;
    power = power >> 1;
  }
  return result;
}
long long modInverse(long long n, long long MOD) {
  return fpow(n, MOD - 2, MOD);
}
int fact[] = {1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int a, b, l = n - i + 1;
    for (int k = 1; k <= m; k++) {
      a = 1, b = l;
      for (int j = 0; j < k - 1; j++) {
        a = (1LL * a * (i + j)) % MD;
        b = (1LL * b * (l + j + 1)) % MD;
      }
      a = (1LL * a * modInverse(fact[k - 1], MD)) % MD;
      b = (1LL * b * modInverse(fact[k], MD)) % MD;
    }
    ans = (ans + (1LL * a * b) % MD) % MD;
  }
  cout << ans << "\n";
  return 0;
}
