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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void __print(int32_t x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
char myHeap[200 * 1024 * 1024];
long long sz = 0;
void assert_tle(bool q) {
  while (!q)
    ;
}
void *operator new(std::size_t count) {
  sz += count;
  assert_tle(sz <= 200 * 1024 * 1024);
  return myHeap + sz - count;
}
void operator delete(void *ptr) {}
void fastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long power(long long a, long long b) {
  long long ans = 1;
  while (b != 0) {
    if (b & 1) {
      ans *= a;
      ans %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    b >>= 1;
  }
  return ans;
}
long long factorial(long long n) {
  if (n == 0) return 1;
  long long ans = n * factorial(n - 1);
  ans %= 1000000007;
  return ans;
}
long long ifactorial(long long n) {
  long long k = factorial(n);
  return power(k, 1000000007 - 2);
}
void solve() {
  long long n, m;
  cin >> n >> m;
  ;
  long long ans = factorial(n + 2 * m - 1) * (ifactorial(2 * m));
  ans %= 1000000007;
  ans *= ifactorial(n - 1);
  ans %= 1000000007;
  cout << ans;
}
int32_t main() {
  fastIO();
  {
    solve();
    cout << "\n";
  }
  return 0;
}
