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;
template <typename T>
inline T MIN(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline T MAX(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T, typename... Args>
inline T MIN(const T &a, const T &b, Args... args) {
  return MIN(MIN(a, b), args...);
}
template <typename T, typename... Args>
inline T MAX(const T &a, const T &b, Args... args) {
  return MAX(MAX(a, b), args...);
}
const double eps = 1e-8;
const int INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int dir[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
int dp[1010];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) dp[i] = i;
  for (int i = 2; i <= 2 * m; i++)
    for (int j = 1; j <= n; j++) dp[j] = (dp[j - 1] + dp[j]) % mod;
  cout << dp[n] << '\n';
  return 0;
}
