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 MAX_N = 1e5 + 5, MOD = 1e9 + 7;
long long factorial[MAX_N + 5], inverse_factorial[MAX_N + 5];
long long power(long long x, long long p) {
  long long result = 1;
  while (p) {
    if (p % 2 == 1) result = (result * x) % MOD;
    x = (x * x) % MOD;
    p /= 2;
  }
  return result;
}
void precompute() {
  factorial[0] = 1;
  for (int i = 1; i <= MAX_N; i++) {
    factorial[i] = (i * factorial[i - 1]) % MOD;
  }
  inverse_factorial[MAX_N] = power(factorial[MAX_N], MOD - 2);
  for (int i = MAX_N - 1; i >= 0; i--) {
    inverse_factorial[i] = ((i + 1) * inverse_factorial[i + 1]) % MOD;
  }
}
long long combinations(long long n, long long k) {
  if (n < k) {
    return 0;
  }
  long long numerator = factorial[n];
  long long denominator_inverse =
      (inverse_factorial[k] * inverse_factorial[n - k]) % MOD;
  return (numerator * denominator_inverse) % MOD;
}
long long count(long long left, long long right, long long length) {
  long long stars = length - 1;
  long long bars = right - left;
  return combinations(stars + bars, stars);
}
int main() {
  precompute();
  int n, length;
  cin >> n >> length;
  long long answer = 0;
  for (int a_ending = 1; a_ending <= n; a_ending++) {
    for (int b_ending = a_ending; b_ending <= n; b_ending++) {
      answer += (count(1, a_ending, length) * count(b_ending, n, length)) % MOD;
      answer %= MOD;
    }
  }
  cout << answer << "\n";
  return 0;
}
