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 MOD = 1e9 + 7;
const int MAX = 1050;
long long fact[MAX];
void calculate() {
  fact[0] = 1;
  for (int i = 1; i < MAX; i++) {
    fact[i] = (fact[i - 1] * i) % MOD;
  }
}
long long modExp(long long b, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) {
      res = (res * b) % MOD;
    }
    b = (b * b) % MOD;
    p >>= 1;
  }
  return res;
}
long long modInverse(long long n) { return modExp(n, MOD - 2); }
long long nCr(long long n, long long r) {
  return (fact[n] * modInverse(fact[n - r]) % MOD * modInverse(fact[r]) % MOD) %
         MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  calculate();
  int n, m;
  cin >> n >> m;
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = (ans + nCr(m + n - i, n - i) * nCr(m + i - 2, i - 1)) % MOD;
  }
  cout << ans;
  return 0;
}
