Greg has a pad. The pad's screen is an n × m rectangle, each cell can be either black or white. We'll consider the pad rows to be numbered with integers from 1 to n from top to bottom. Similarly, the pad's columns are numbered with integers from 1 to m from left to right.

Greg thinks that the pad's screen displays a cave if the following conditions hold:

  * There is a segment [l, r] (1 ≤ l ≤ r ≤ n), such that each of the rows l, l + 1, ..., r has exactly two black cells and all other rows have only white cells. 
  * There is a row number t (l ≤ t ≤ r), such that for all pairs of rows with numbers i and j (l ≤ i ≤ j ≤ t) the set of columns between the black cells in row i (with the columns where is these black cells) is the subset of the set of columns between the black cells in row j (with the columns where is these black cells). Similarly, for all pairs of rows with numbers i and j (t ≤ i ≤ j ≤ r) the set of columns between the black cells in row j (with the columns where is these black cells) is the subset of the set of columns between the black cells in row i (with the columns where is these black cells). 



Greg wondered, how many ways there are to paint a cave on his pad. Two ways can be considered distinct if there is a cell that has distinct colors on the two pictures.

Help Greg.

Input

The first line contains two integers n, m — the pad's screen size (1 ≤ n, m ≤ 2000).

Output

In the single line print the remainder after dividing the answer to the problem by 1000000007 (109 + 7).

Examples

Input

1 1


Output

0


Input

4 4


Output

485


Input

3 5


Output

451
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 2010;
int n, m;
long long f[N][N], g[N][N], h[N][N], o[N][N], p[N][N], ans;
int main() {
  cin >> n >> m;
  memset(f, 0, sizeof(f));
  memset(g, 0, sizeof(g));
  memset(h, 0, sizeof(h));
  memset(o, 0, sizeof(o));
  for (int i = 1; i <= n; ++i) {
    for (int j = 2; j <= m; ++j) {
      (f[i][j] += 1 + g[i - 1][j] * j % mod + h[i - 1][j]) %= mod;
      o[i][j] = (f[i][j] - f[i - 1][j]) % mod;
    }
    for (int j = 2; j <= m; ++j) {
      g[i][j] = (g[i][j - 1] + f[i][j]) % mod;
      h[i][j] = (h[i][j - 1] + f[i][j] * (1 - j) % mod) % mod;
    }
  }
  memset(p, 0, sizeof(p));
  for (int i = 1; i <= n; ++i)
    for (int j = 2; j <= m; ++j) p[i][j] = (p[i - 1][j] + o[i][j]) % mod;
  ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 2; j <= m; ++j) {
      (ans += (m - j + 1) * o[i][j] % mod * p[n - i + 1][j] % mod) %= mod;
    }
  cout << (ans + mod) % mod << endl;
  return 0;
}
