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;
template <class T>
inline T read() {
  T x = 0, f = 1;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  };
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + s - 48;
    s = getchar();
  };
  return x * f;
}
template <typename T>
inline void wri(T x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  };
  if (x / 10) wri(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void print(T x, char s) {
  wri(x);
  putchar(s);
}
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long C[1010][15];
long long cal(int n, int m) {
  if (C[n][m] || n == 0)
    return C[n][m];
  else {
    for (register int i = (0); i < (m + 1); i++) {
      if (i == 0)
        C[n][m] = (C[n][m] + 1) % mod;
      else {
        C[n][m] = (C[n][m] + cal(n - 1, i)) % mod;
      }
    }
  }
  return C[n][m];
}
long long cnt;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (register int i = (1); i < (m + 1); i++) C[1][i] = 1;
  for (register int i = (1); i < (n + 1); i++) {
    cnt =
        ((cnt + ((cal(i, m) - cal(i - 1, m) + mod) % mod * cal(n - i + 1, m))) %
         mod) %
        mod;
  }
  cout << cnt;
  return 0;
}
