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;
int n, m;
int dp[1010][15];
int add(int a, int b) {
  int ret = a + b;
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
int sub(int a, int b) {
  int ret = a - b;
  if (ret < 0) ret += 1000000007;
  return ret;
}
int mul(long long a, long long b) { return (a * b) % 1000000007; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) dp[i][1] = 1;
  for (int j = 1; j <= m; j++) dp[1][j] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 2; j <= m; j++) dp[i][j] = add(dp[i - 1][j], dp[i][j - 1]);
  int sum = 0;
  for (int i = 1; i <= n; i++) sum = add(sum, dp[i][m]);
  int sol = 0;
  for (int i = 1; i <= n; i++)
    sol = add(sol, mul(dp[i][m], sum)), sum = sub(sum, dp[n - i + 1][m]);
  cout << sol;
  return 0;
}
