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 MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
long long fact[MAXN];
long long POW(long long a, long long b) {
  if (b == 0) return 1;
  long long ans = POW(a, b / 2);
  ans = ans * ans % MOD;
  if (b % 2) ans = ans * a % MOD;
  return ans;
}
long long C(long long a, long long b) {
  return fact[a] * (POW(fact[b] * fact[a - b] % MOD, MOD - 2) % MOD) % MOD;
}
long long dp[15][MAXN];
int main() {
  fact[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    fact[i] = i * fact[i - 1] % MOD;
  }
  int n, m;
  scanf("%d%d", &n, &m);
  long long ans = C(n + 2 * m - 1, 2 * m);
  printf("%lld\n", ans);
}
