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 int64_t inf = 1e9 + 7;
int main() {
  int64_t n, m;
  cin >> n >> m;
  vector<vector<int64_t>> cnta = vector<vector<int64_t>>(m, vector<int64_t>(n));
  for (int64_t i = 0; i < n; ++i) {
    cnta[0][i] = 1;
  }
  for (int64_t i = 1; i < m; ++i) {
    for (int64_t j = 0; j < n; ++j) {
      for (int64_t k = 0; k <= j; ++k) {
        cnta[i][j] += cnta[i - 1][k];
        cnta[i][j] %= inf;
      }
    }
  }
  vector<vector<int64_t>> cntb = vector<vector<int64_t>>(m, vector<int64_t>(n));
  for (int64_t i = 0; i < n; ++i) {
    cntb[0][i] = 1;
  }
  for (int64_t i = 1; i < m; ++i) {
    for (int64_t j = 0; j < n; ++j) {
      for (int64_t k = j; k < n; ++k) {
        cntb[i][j] += cntb[i - 1][k];
        cntb[i][j] %= inf;
      }
    }
  }
  int64_t cnt = 0;
  vector<int64_t> a = cnta[m - 1];
  vector<int64_t> b = cntb[m - 1];
  for (int64_t i = 0; i < n; ++i) {
    for (int64_t j = i; j < n; ++j) {
      cnt += a[i] * b[j];
      cnt %= inf;
    }
  }
  cout << cnt;
}
