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 maxn = 2 * 1000 + 10, maxm = (1 << 17), mod = (int)1e9 + 7,
          hash = 701, maxim = (1 << 24) + 10, inf = (1 << 29) + 3;
const double pi = 3.14159265359, ee = 2.71828;
long long d[2][maxn][maxn], dp[2][maxn][maxn], ans;
void update(int x, int p, int ind, long long val) {
  for (int i = ind; i < maxn; i += (i & -i)) d[p][x][i] += val;
}
long long query(int x, int p, int ind) {
  long long sum = 0;
  for (int i = ind; i; i -= (i & -i)) sum += d[p][x][i];
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++) {
      dp[1][i][j] = (query(i - 1, 0, j) * j - query(i - 1, 1, j) + 1) % mod,
      update(i, 0, j, dp[1][i][j]),
      update(i, 1, j, (dp[1][i][j] * (j - 1)) % mod);
      dp[0][i][j] =
          (query(i - 1, 0, j - 1) * j - query(i - 1, 1, j - 1) + 1) % mod;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      ans += dp[0][i][j] * dp[1][n - i + 1][j] * (m - j + 1), ans %= mod;
  cout << ans << endl;
  return 0;
}
