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 long long INF = 1e17;
const long long N = 2e6 + 77;
const long long MOD = 1e9 + 7;
using namespace std;
vector<long long> prime;
long long m, k, a[N];
long long dp[1105][34];
long long ncr(long long n, long long r) {
  if (n == r) return 1;
  if (r == 0) return 1;
  if (dp[n][r] != -1) return dp[n][r];
  return dp[n][r] = (ncr(n - 1, r - 1) % MOD + ncr(n - 1, r) % MOD) % MOD;
}
long long POWER(long long base, long long expo) {
  long long ret = 1;
  while (expo) {
    if (expo & 1) ret = ret * base;
    expo >>= 1;
    base = base * base;
  }
  return ret;
}
int main() {
  long long test_case = 1;
  while (test_case--) {
    long long nn;
    cin >> nn >> m;
    m *= 2;
    for (long long i = 0; i <= 1100; i++) {
      for (long long j = 0; j <= 30; j++) dp[i][j] = -1;
    }
    cout << ncr(nn + m - 1, m) % MOD;
  }
  return 0;
}
