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 int N = 105;
const int M = 11;
const int md = 1e9 + 7;
int n, m;
int add(int a, int b) {
  a += b;
  if (a >= md) {
    a -= md;
  }
  return a;
}
int sub(int a, int b) { return (a - b + md) % md; }
int mul(int a, int b) { return (1LL * (a % md) * (b % md)) % md; }
int power(int a, int b) {
  int ret = 1;
  while (b) {
    if (b & 1) ret = mul(ret, a);
    a = mul(a, a);
    b /= 2;
  }
  return ret;
}
int invmd(int x) { return power(x, md - 2); }
int fct[2005];
int invf[2005];
int ncr(int n, int r) { return mul(fct[n], mul(invf[n - r], invf[r])); }
void pre() {
  fct[0] = fct[1] = 1;
  invf[0] = invf[1] = 1;
  for (int i = 2; i < 2005; ++i) {
    fct[i] = mul(fct[i - 1], i);
    invf[i] = invmd(fct[i]);
  }
}
int C(int n, int r) { return ncr(n, r); }
void getAC() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  pre();
  int ans = C(2 * m + n - 1, n - 1);
  cout << ans << '\n';
}
int main() {
  getAC();
  return 0;
}
