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 long long mod = 1e9 + 7;
const long long maxn = 1e5 + 5;
long long n, m, ans = 0;
long long c[maxn], jc[maxn];
long long ksm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
long long niyuan(long long k) { return ksm(k, mod - 2) % mod; }
long long C(long long n, long long m) {
  if (m > n) return 0;
  if (m == 0 || n == m)
    return 1;
  else
    return ((jc[n] * niyuan(jc[m])) % mod * niyuan(jc[n - m])) % mod;
}
void init() {
  jc[0] = 1;
  for (long long i = 1; i <= 100000; i++) jc[i] = (jc[i - 1] * i) % mod;
  for (long long i = 1; i <= 1000; i++) c[i] = C(i + (m - 1) - 1, m - 1);
}
signed main() {
  scanf("%lld %lld", &n, &m);
  init();
  for (long long i = 1; i <= n; i++) {
    for (long long j = i; j <= n; j++)
      ans = (ans + (c[i] * c[n - j + 1]) % mod) % mod;
  }
  printf("%lld", ans);
  return 0;
}
