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 INF = (long long)1e18 + 7, t = -1, till = 0, N = 1e9 + 7;
bool comp(pair<long long, long long> p1, pair<long long, long long> p2) {
  return p1.first - p1.second > p2.first - p2.second;
}
long long ss(long long x, long long y) { return (x - y) * (x - y); }
void print(long long f) {
  if (f)
    cout << "YES" << '\n';
  else
    cout << "NO" << '\n';
}
long long prime(long long a) {
  if (a == 1) return 0;
  for (long long i = 2; i <= round(sqrt(a)); ++i) {
    if (a % i == 0) {
      return 0;
    }
  }
  return 1;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long dp[2 * m + 1][n + 1];
  for (long long i = 0; i <= n; i++) {
    dp[0][i] = 1;
  }
  long long ans = 0;
  for (long long len = 1; len <= 2 * m; len++) {
    long long s = 0;
    for (long long digit = 1; digit <= n; digit++) {
      dp[len][digit] = dp[len - 1][digit];
    }
    for (long long digit = 1; digit <= n; digit++) {
      s += dp[len][digit];
      s %= N;
      dp[len][digit] = s;
    }
    if (len == 2 * m) ans = s;
  }
  cout << ans << '\n';
}
