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;
int const MOD = (int)1e9 + 7;
inline void normal(long long &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long modMul(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long modAdd(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long modSub(long long a, long long b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long modPow(long long b, long long p) {
  long long r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long modInverse(long long a) { return modPow(a, MOD - 2); }
inline long long modDiv(long long a, long long b) {
  return modMul(a, modInverse(b));
}
inline long long mod(long long a, long long m) {
  if (a < 0)
    return m - (abs(a) % m);
  else if (a < m)
    return a;
  else
    return (a % m);
}
long long nCr(long long n, long long r) {
  if (r > n - r) r - n - r;
  long long ans = 1;
  for (int i = 1; i <= r; i++) {
    ans = modMul(ans, n - r + i);
    ans = modDiv(ans, i);
  }
  return ans % MOD;
}
int main() {
  long long n, m;
  cin >> n >> m;
  cout << nCr(n + 2 * m - 1, 2 * m) << endl;
}
