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;
using namespace std::chrono;
using ii = pair<long long int, long long int>;
class prioritize {
 public:
  bool operator()(ii &p1, ii &p2) { return p1.first < p2.first; }
};
auto start_time = high_resolution_clock::now();
void time() { return; }
void ojudge() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  return;
}
long long int fact[1030];
long long int powerm(long long int a, long long int b, long long int m) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a + m) % m;
    a = (a * a + m) % m;
    b >>= 1;
  }
  return ans;
}
long long int invmod(long long int a, long long int m) {
  return (powerm(a, m - 2, m) + m) % m;
}
void fact_mod(long long int n, long long int m) {
  fact[0] = 1;
  for (long long int i = 1; i < n + 1; i++) {
    fact[i] = (i * fact[i - 1]) % m;
  }
}
long long int nCr(long long int n, long long int r, long long int m) {
  fact_mod(n, m);
  long long int denom = (fact[r] * fact[n - r] + m) % m;
  denom = invmod(denom, m);
  return (fact[n] * denom + m) % m;
}
int main() {
  ojudge();
  long long int n, m;
  cin >> n >> m;
  long long int count = nCr(2 * m + n - 1, n - 1, 1000000007);
  cout << count << "\n";
  time();
  return 0;
}
