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 MAX_N = 2000 + 10;
const int MOD = 1000000000 + 7;
int N, M;
long long f[MAX_N][MAX_N], g[MAX_N][MAX_N], c[MAX_N][MAX_N];
long long l1[MAX_N][MAX_N], l2[MAX_N][MAX_N];
void solve() {
  cin >> N >> M;
  for (int j = 0; j <= M; ++j) g[0][j] = 1;
  for (int i = 1; i <= N; ++i)
    for (int j = 2; j <= M; ++j) {
      c[i][j] = g[i - 1][j];
      f[i][j] = (f[i][j - 1] + c[i][j]) % MOD;
      g[i][j] = (g[i][j - 1] + f[i][j]) % MOD;
    }
  for (int j = 2; j <= M; ++j)
    for (int i = 1; i <= N; ++i) {
      l1[i][j] = (l1[i - 1][j] + c[i][j]) % MOD;
      l2[i][j] = (l2[i - 1][j] + g[i][j]) % MOD;
    }
  long long ret = 0;
  for (int i = 1; i <= N; ++i)
    for (int j = 2; j <= M; ++j)
      ret = (ret + l1[i][j] * (l2[N - i][j] + 1 - l1[N - i][j] + MOD) % MOD *
                       (M - j + 1)) %
            MOD;
  cout << ret << endl;
}
int main() {
  solve();
  return 0;
}
