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 = 2010;
const int MOD = 1000 * 1000 * 1000 + 7;
long long int dp[MAXN][MAXN];
long long int sum[MAXN][MAXN];
long long int sum2[MAXN][MAXN];
void M(long long int &a) { a = a % MOD; }
int main() {
  for (int i = 2; i < MAXN; i++) dp[0][i] = 1;
  for (int h = 0; h < MAXN - 1; h++) {
    long long int s = 0, s2 = 0;
    for (int l = 2; l < MAXN; l++) {
      s += s2;
      s += dp[h][l];
      s2 += dp[h][l];
      M(s);
      M(s2);
      dp[h + 1][l] = s;
    }
  }
  int n, m;
  cin >> n >> m;
  for (int i = 2; i <= m; i++)
    for (int j = 0; j < n; j++) {
      if (j) sum[j][i] = sum[j - 1][i];
      sum[j][i] += dp[j][i];
      M(sum[j][i]);
    }
  for (int i = 0; i < n; i++) {
    long long int s = 0, s2 = 0;
    for (int j = 2; j <= m; j++) {
      sum2[i + 1][j] = s;
      s += sum[i][j];
      s2 += sum[i][j];
      s += s2;
      M(s);
      M(s2);
    }
  }
  long long int ans = 0;
  for (int i = 2; i <= m; i++)
    for (int j = 0; j < n; j++) {
      long long int t = sum[j][i] * (m - i + 1);
      M(t);
      t *= sum2[n - j - 1][i] + 1;
      M(t);
      ans += t;
      M(ans);
    }
  cout << ans << endl;
  return 0;
}
