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 = (long long)1e9 + 7;
long long n, m;
long long dp[1010][1010][15];
long long fn(long long i, long long j, long long k) {
  if (i > j) return 0;
  if (k >= m) return 1;
  if (dp[i][j][k] != -1) return dp[i][j][k];
  long long ans = 0;
  ans += fn(i + 1, j, k);
  ans %= mod;
  ans += fn(i, j - 1, k);
  ans %= mod;
  ans += fn(i, j, k + 1);
  ans %= mod;
  ans -= fn(i + 1, j - 1, k);
  return dp[i][j][k] = ans % mod;
}
void solve() {
  cin >> n >> m;
  memset(dp, -1, sizeof(dp));
  ;
  long long value = fn(1, n, 0);
  if (value < 0)
    cout << value + mod;
  else
    cout << value;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
}
