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 md = 1e9 + 7;
int add(int a, int b) {
  a += b;
  if (a >= md) {
    a -= md;
  }
  return a;
}
int solve(int pos, int last, int& m, int& n, vector<vector<int>>& dp) {
  if (pos == m) {
    return 1;
  }
  if (dp[pos][last] != -1) return dp[pos][last];
  int res = 0;
  for (int cur = max(1, last); cur <= n; ++cur) {
    res = add(res, solve(pos + 1, cur, m, n, dp));
  }
  return dp[pos][last] = res;
}
void getAC() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  m <<= 1;
  vector<vector<int>> dp(m, vector<int>(n + 1, -1));
  cout << solve(0, 0, m, n, dp) << '\n';
}
int main() {
  getAC();
  return 0;
}
