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 int INF = 0x3f3f3f3f, N = 1e6 + 5, MOD = 1e9 + 7;
const long long INF_LL = 0x3f3f3f3f3f3f3f3fLL;
inline int getplc(int x, int y) { return (x >> y) & 1; }
template <typename T>
T square(T x) {
  return x * x;
}
long long qpow(long long a, long long b = MOD - 2, long long _MOD = MOD) {
  long long res = 1;
  for (; b; b >>= 1, a = a * a % _MOD) {
    if (b & 1) res = res * a % _MOD;
  }
  return res;
}
template <typename T>
T Smax(T x) {
  return x;
}
template <typename T, typename... Args>
T Smax(T a, Args... args) {
  return max(a, Smax(args...));
}
template <typename T>
T Smin(T x) {
  return x;
}
template <typename T, typename... Args>
T Smin(T a, Args... args) {
  return min(a, Smin(args...));
}
template <typename T>
void errl(istream_iterator<string> it) {}
template <typename T, typename... Args>
void errl(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  errl(++it, args...);
}
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << "=" << a << " # ";
  err(++it, args...);
}
void Solve();
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  Solve();
  return 0;
}
int fac[N], inv[N];
void init(int n) {
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; i++) fac[i] = (long long)fac[i - 1] * i % MOD;
  for (int i = 1; i <= n; i++) inv[i] = qpow(fac[i], MOD - 2, MOD);
}
int Comb(int n, int m) {
  if (n - m >= 0 && m >= 0)
    return 1ll * fac[n] * inv[n - m] % MOD * inv[m] % MOD;
  else
    return 0;
}
void Solve() {
  int n, m;
  cin >> n >> m;
  init(max(n, m) * 2);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= i; j++) {
      ans = (ans +
             (long long)Comb(m + j - 2, m - 1) * Comb(m + n - i - 1, m - 1)) %
            MOD;
    }
  }
  cout << ans << endl;
}
