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 modd = 1000000000 + 7;
const int maxn = 2000 + 15;
int f[maxn][maxn], g[maxn][maxn], sum[maxn][maxn];
int n, m;
long long ans;
int main() {
  scanf("%d%d", &n, &m);
  if (m == 1) {
    printf("0\n");
    return 0;
  }
  for (int i = 0; i <= m - 2; i++) f[1][i] = 1;
  for (int i = 2; i <= n; i++)
    for (int j = 0; j <= m - 2; j++) {
      if (j == 0)
        g[i][j] = f[i - 1][j], f[i][j] = g[i][j];
      else {
        g[i][j] = g[i][j - 1] + f[i - 1][j];
        if (g[i][j] >= modd) g[i][j] -= modd;
        f[i][j] = f[i][j - 1] + g[i][j];
        if (f[i][j] >= modd) f[i][j] -= modd;
      }
    }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= m - 2; j++) {
      sum[i][j] = sum[i - 1][j] + f[i][j];
      if (sum[i][j] >= modd) sum[i][j] -= modd;
    }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= m - 2; j++) {
      ans += 1ll * (sum[i][j] - sum[i - 1][j]) * sum[n - i + 1][j] % modd *
             (m - 2 - j + 1) % modd;
      if (ans >= modd) ans -= modd;
      if (ans < 0) ans += modd;
    }
  printf("%I64d\n", ans);
  return 0;
}
