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;
long long int max(long long int a, long long int b) {
  if (a > b) {
    return a;
  }
  return b;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long int min(long long int a, long long int b) {
  if (a < b) {
    return a;
  }
  return b;
}
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      res = (res * x);
    }
    x = (x * x);
    n = n / 2;
  }
  return res;
}
long long int factorialNumInverse[1000001];
long long int naturalNumInverse[1000001];
long long int fact[1000001];
void InverseofNumber(long long int n, long long int p) {
  naturalNumInverse[0] = naturalNumInverse[1] = 1;
  for (long long int i = 2; i <= n; i++)
    naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
}
void InverseofFactorial(long long int n, long long int p) {
  factorialNumInverse[0] = factorialNumInverse[1] = 1;
  for (long long int i = 2; i <= n; i++)
    factorialNumInverse[i] =
        (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
}
void factorial(long long int n, long long int p) {
  fact[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    fact[i] = (fact[i - 1] * i) % p;
  }
}
long long int nCrModp(long long int n, long long int r, long long int p) {
  long long int ans =
      ((fact[n] * factorialNumInverse[r]) % p * factorialNumInverse[n - r]) % p;
  return ans;
}
long long int n, m;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  factorial(200005, 1000000007);
  InverseofNumber(200005, 1000000007);
  InverseofFactorial(200005, 1000000007);
  cin >> n >> m;
  cout << nCrModp(n + 2 * m - 1, 2 * m, 1000000007) << "\n";
  return 0;
}
