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>
#pragma warning(disable : 4996)
const long long MOD = 1000000007;
using namespace std;
const int max_comb = 30000;
vector<long long> fac(max_comb + 1);
vector<long long> ifac(max_comb + 1);
long long mpow(long long x, long long n) {
  long long ans = 1;
  while (n != 0) {
    if (n & 1) ans = ans * x % MOD;
    x = x * x % MOD;
    n = n >> 1;
  }
  return ans;
}
long long minv(long long x) { return mpow(x, MOD - 2); }
long long comb(int a, int b) {
  if (a == 0 && b == 0) return 1;
  if (a < b || a < 0) return 0;
  long long tmp = ifac[a - b] * ifac[b] % MOD;
  return tmp * fac[a] % MOD;
}
long long perm(int a, int b) {
  if (b == 0) return 1;
  if (a < b || a < 0) return 0;
  long long tmp = ifac[a - b] % MOD;
  return tmp * fac[a] % MOD;
}
void pre_comb() {
  fac[0] = 1;
  ifac[0] = 1;
  for (int i = 0; i < max_comb; i++) {
    fac[i + 1] = fac[i] * (i + 1) % MOD;
    ifac[i + 1] = ifac[i] * minv(i + 1) % MOD;
  }
  return;
}
int main(int argc, char* argv[]) {
  int n, m;
  scanf("%d%d", &n, &m);
  pre_comb();
  long long ans = comb(n - 1 + m * 2, n - 1);
  printf("%lld\n", ans);
  return 0;
}
