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>
const double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
using namespace std;
const unsigned long long int MOD = 1e9 + 7;
double gcd(double a, double b) { return a < 0.01 ? b : gcd(fmod(b, a), a); }
template <typename T>
T mymax(T a, T b) {
  return (a > b) ? a : b;
}
template <typename T>
T mymax(T a, T b, T c) {
  return mymax(a, mymax(b, c));
}
template <typename T>
T mymin(T a, T b) {
  return (a < b) ? a : b;
}
template <typename T>
T mymin(T a, T b, T c) {
  return mymin(a, mymin(b, c));
}
template <typename T>
T power(T x, T y) {
  T res = 1;
  x = x % MOD;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1;
    x = (x * x) % MOD;
  }
  return res;
}
template <typename T>
void swap(T *x, T *y) {
  T temp;
  temp = *y;
  *y = *x;
  *x = temp;
}
template <typename T>
T mod(T a) {
  if (a > 0)
    return a;
  else
    return -a;
}
template <typename T>
T logg(T a) {
  T x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
unsigned long long int ans = 0;
unsigned long long int n, m;
unsigned long long int pw[20];
unsigned long long int countNonDecreasing(unsigned long long int n,
                                          unsigned long long int a) {
  unsigned long long int N = a;
  unsigned long long int count = 1;
  for (unsigned long long int i = 1; i <= n; i++) {
    count *= (N + i - 1);
    count %= MOD;
    count *= pw[i];
    count %= MOD;
  }
  return count;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m;
  for (unsigned long long int i = 1; i <= m; i++) {
    pw[i] = power(i, MOD - 2);
    pw[i] %= MOD;
  }
  for (unsigned long long int i = 1; i <= n; i++) {
    unsigned long long int k = countNonDecreasing(m - 1, i);
    for (unsigned long long int j = i; j <= n; j++) {
      unsigned long long int p = (k);
      p *= countNonDecreasing(m - 1, n - j + 1);
      p %= MOD;
      ans += p;
      ans %= MOD;
    }
  }
  cout << ans << '\n';
  return 0;
}
