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;
auto clk = clock();
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long INF = 1e18L + 5;
const int d4i[4] = {-1, 0, 1, 0}, d4j[4] = {0, 1, 0, -1};
const int d8i[8] = {-1, -1, 0, 1, 1, 1, 0, -1},
          d8j[8] = {0, 1, 1, 1, 0, -1, -1, -1};
long long modexp(long long base, long long power) {
  if (base == 0) return 0ll;
  if (power == 0) return 1ll;
  long long res = 1;
  while (power) {
    if (power & 1) res = (res * base) % 1000000007;
    base = (base * base) % 1000000007;
    power /= 2;
  }
  return res;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  long long x = n + 2 * m - 1;
  long long f[x + 1];
  f[0] = 1;
  for (long long i = (1), _b = (x); i <= _b; i++) {
    f[i] = (f[i - 1] * i) % 1000000007;
  }
  long long ans = f[x];
  ans = (ans * (modexp(f[2 * m], 1000000007 - 2))) % 1000000007;
  ans = (ans * (modexp(f[n - 1], 1000000007 - 2))) % 1000000007;
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(11);
  long long t = 1;
  while (t--) solve();
  return 0;
}
