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;
long long dp2[2010][2010];
long long dp1[2010][2010];
int N, M;
long long sums1[2010][2010], sums[2010][2010];
inline void go(void) {
  for (int i = 2; i <= M; i++) {
    dp2[1][i] = 1;
    sums1[1][i] = sums1[1][i - 1] + (1LL * (M - i + 1)) % 1000000007;
    if (sums1[1][i] >= 1000000007) sums1[1][i] -= 1000000007;
    sums[1][i] = sums[1][i - 1] + 1LL;
    if (sums[1][i] >= 1000000007) sums[1][i] -= 1000000007;
    dp1[1][i] = sums1[1][i - 1] - (sums[1][i - 1] * (M - i)) % 1000000007;
    if (dp1[1][i] < 0) dp1[1][i] += 1000000007;
  }
  for (int i = 2; i <= N; i++) {
    for (int j = 2; j <= M; j++) {
      dp2[i][j] = sums1[i - 1][j] - (sums[i - 1][j] * (M - j)) % 1000000007;
      if (dp2[i][j] < 0) dp2[i][j] += 1000000007;
      sums1[i][j] = sums1[i][j - 1] + (dp2[i][j] * (M - j + 1)) % 1000000007;
      if (sums1[i][j] >= 1000000007) sums1[i][j] -= 1000000007;
      sums[i][j] = sums[i][j - 1] + dp2[i][j];
      if (sums[i][j] >= 1000000007) sums[i][j] -= 1000000007;
      dp2[i][j] += dp2[i - 1][j];
      if (dp2[i][j] >= 1000000007) dp2[i][j] -= 1000000007;
      dp1[i][j] = sums1[i][j - 1] - (sums[i][j - 1] * (M - j)) % 1000000007;
      if (dp1[i][j] < 0) dp1[i][j] += 1000000007;
      dp1[i][j] += dp1[i - 1][j];
      if (dp1[i][j] >= 1000000007) dp1[i][j] -= 1000000007;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= N; i++) {
    int u, v;
    u = i - 1;
    v = N - i + 1;
    for (int t = 2; t <= M; t++) {
      long long foo = (dp1[u][t] * dp2[v][t]) % 1000000007;
      foo *= (M - t + 1);
      foo %= 1000000007;
      ans += foo;
      ans %= 1000000007;
      foo = (dp2[i][t] * (M - t + 1)) % 1000000007;
      foo %= 1000000007;
      ans += foo;
      if (ans >= 1000000007) ans -= 1000000007;
    }
  }
  cout << ans << endl;
}
inline void ReadInput(void) {
  scanf("%d", &N);
  scanf("%d", &M);
}
inline void solve(void) { go(); }
inline void Refresh(void) {}
int main() {
  ReadInput();
  solve();
  return 0;
}
