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 = 2e3 + 6.66;
const int MAXM = 100 + 6.66;
const long long MOD = 1e9 + 7;
long long dp[MAXN][MAXM];
long long dp1[MAXN][MAXM];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < m - 1; j++) {
      if (i == 1) {
        dp[i][j] = 1;
        dp1[i][j] = 1;
        continue;
      }
      for (int k = 0; k < j; k++) {
        dp[i][j] = (dp[i][j] + 1LL * dp[i - 1][k] * (j - k + 1LL) % MOD) % MOD;
      }
      dp1[i][j] = dp[i][j] % MOD;
      dp[i][j] = (dp[i][j] + 1LL * dp[i - 1][j] % MOD) % MOD;
    }
  }
  long long ans = 0LL;
  for (int l = 0; l < n; l++)
    for (int r = 0; r < n - l; r++) {
      for (int sz = 0; sz < m - 1; sz++) {
        long long WW = (dp1[l + 1][sz] * 1LL * dp[r + 1][sz]) % MOD;
        if (WW < 0) cout << "[ !" << 53 << "L ] " << endl;
        ;
        ans = (ans +
               1LL * (n - (r + l + 1) + 1) * ((m - sz - 1) * 1LL * WW) % MOD) %
              MOD;
      }
    }
  cout << ans << endl;
  return 0;
}
