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;
long long call(int i, int n, vector<vector<long long> >& dp,
               vector<long long>& v) {
  if (i == -1) {
    return 1;
  }
  if (dp[i][n] != -1) {
    return dp[i][n];
  }
  dp[i][n] = 0;
  for (int j = 1; j <= n; j++) {
    long long p = call(i - 1, j, dp, v);
    dp[i][n] = (dp[i][n] + p) % 1000000007;
    if (i == dp.size() - 1) {
      v[j] = p;
    }
  }
  return dp[i][n];
}
long long call1(int i, int n, vector<vector<long long> >& dp,
                vector<long long>& v1) {
  if (i == -1) {
    return 1;
  }
  if (dp[i][n] != -1) {
    return dp[i][n];
  }
  dp[i][n] = 0;
  for (int j = n; j < dp[0].size(); j++) {
    long long p = call1(i - 1, j, dp, v1);
    dp[i][n] = (dp[i][n] + p) % 1000000007;
    if (i == dp.size() - 1) {
      v1[j] = p;
    }
  }
  return dp[i][n];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<long long> v(n + 1);
  vector<long long> v1(n + 1);
  vector<vector<long long> > dp(m, vector<long long>(n + 1, -1));
  call(m - 1, n, dp, v);
  vector<vector<long long> > dp1(m, vector<long long>(n + 1, -1));
  call1(m - 1, 1, dp1, v1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 1; j--) {
      ans = (ans + (v1[i] * v[j]) % 1000000007) % 1000000007;
    }
  }
  cout << ans;
}
