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 = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
long long fac[2005];
long long qp(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
long long inv(long long x) { return qp(x, mod - 2); }
long long c(long long m, long long n) {
  return fac[n] * inv(fac[m]) % mod * inv(fac[n - m]) % mod;
}
int n, m;
void solve() { printf("%lld\n", c(2 * m, n + 2 * m - 1)); }
int main() {
  fac[0] = 1;
  for (int i = 1; i <= 2005; i++) fac[i] = fac[i - 1] * i % mod;
  while (~scanf("%d %d", &n, &m)) solve();
  return 0;
}
