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;
int n, m;
const int MAX = 2010;
const int MOD = 1000000000 + 7;
int ways[MAX][MAX], ways2[MAX][MAX];
int main() {
  scanf("%d %d", &n, &m);
  if (m < 2) {
    puts("0");
    return 0;
  }
  --m;
  for (int i = 1; i <= m; i++) ways[1][i] = ways2[1][i] = 1;
  for (int i = 2; i <= n; i++) {
    long long sum = 0, add = 0;
    for (int j = 1; j <= m; j++) {
      ways2[i][j] = (sum * j + MOD - add) % MOD;
      sum = (sum + ways[i - 1][j]) % MOD;
      add = (add + (long long)ways[i - 1][j] * (j - 1)) % MOD;
      ways[i][j] = (sum * j + MOD - add) % MOD;
    }
  }
  int ans = 0;
  for (int i = 1; i <= m; i++) {
    long long cur = 0, sum = 0, add = 0;
    for (int j = n; j >= 1; j--) {
      int k = n - j + 1;
      add = (add + (long long)ways2[k][i] * (n + 2 - k)) % MOD;
      sum = (sum + ways2[k][i]) % MOD;
      cur = (cur + (long long)ways[j][i] * (add + MOD - sum * j % MOD)) % MOD;
    }
    ans = (ans + cur * (m - i + 1)) % MOD;
  }
  printf("%d\n", ans);
}
