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 int mod = int(1e9 + 7);
int n, m, first[2222][2222], g[2222][2222];
int main() {
  scanf("%d%d", &n, &m);
  for (int w = 2; w <= m; ++w) first[1][w] = g[1][w] = 1;
  for (int h = 2; h <= n; ++h) {
    int sum = 0, total = 0;
    for (int w = 2; w <= m; ++w) {
      (sum += first[h - 1][w]) %= mod;
      total = (1LL * first[h - 1][w] * w + total) % mod;
      first[h][w] = (1LL * sum * (w + 1) - total + mod) % mod;
    }
    sum = 0, total = 0;
    for (int w = 2; w <= m; ++w) {
      (sum += g[h - 1][w]) %= mod;
      total = (1LL * g[h - 1][w] * w + total) % mod;
      g[h][w] = (1LL * sum * (w + 1) - total + mod + h) % mod;
    }
  }
  int res = 0;
  for (int h = 1; h <= n; ++h) {
    int r = n - h + 1;
    int sum = 0, total = 0;
    for (int w = 2; w <= m; ++w) {
      int cur = (1LL * sum * (w + 1) - total + mod + r) % mod;
      cur = (1LL * cur * (m - w + 1)) % mod;
      res = (1LL * cur * first[h][w] + res) % mod;
      (sum += g[r - 1][w]) %= mod;
      total = (1LL * g[r - 1][w] * w + total) % mod;
    }
  }
  cout << res << endl;
  return 0;
}
