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;
template <typename T, typename U>
inline void amin(T &a, U b) {
  a = (a > b ? b : a);
}
template <typename T, typename U>
inline void amax(T &a, U b) {
  a = (a > b ? a : b);
}
const int N = 2e3 + 5, mod = 1e9 + 7;
int64_t fac[N], inv[N];
long long int expo(long long int n, long long int p, long long int M) {
  long long int res = 1;
  while (p) {
    if (p & 1) res = (res * n) % M;
    n = (n * n) % M;
    p >>= 1;
  }
  return res;
}
void pre() {
  fac[0] = 1;
  inv[0] = 1;
  for (int i = 1; i < N; ++i) {
    fac[i] = fac[i - 1] * i;
    fac[i] %= mod;
    inv[i] = expo(fac[i], mod - 2, mod);
  }
}
int64_t nck(int n, int k) { return (fac[n] * inv[n - k] % mod * inv[k] % mod); }
void solve() {
  int n, m;
  cin >> n >> m;
  cout << nck(n + 2 * m - 1, n - 1) << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  pre();
  while (t--) {
    solve();
  }
}
