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 xd = 1e6;
const long long mod = 1e9 + 7;
double dot(complex<double> a, complex<double> b) {
  return (conj(a) * b).real();
}
double cross(complex<double> a, complex<double> b) {
  return (conj(a) * b).imag();
}
long long fact[1000006], inv[1000006];
long long fp(long long base, long long exp) {
  if (exp == 0) return 1;
  long long ans = fp(base, exp / 2);
  ans = (ans * ans) % mod;
  if (exp % 2 != 0) ans = (ans * (base % mod)) % mod;
  return ans;
}
void calcFacAndInv(long long n) {
  fact[0] = inv[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (i * fact[i - 1]) % mod;
    inv[i] = fp(fact[i], mod - 2);
  }
}
long long ncr(long long n, long long r) {
  return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;
}
long long npr(long long n, long long r) { return (fact[n] * inv[n - r]) % mod; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  calcFacAndInv(100000);
  int n, m;
  long long x, ans;
  cin >> n >> m;
  m *= 2;
  x = m + n - 1;
  ans = ncr(x, m) % mod;
  cout << ans;
  return 0;
}
