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 long long MOD = 1000000007;
int main() {
  ios_base::sync_with_stdio(0);
  long long n, m;
  cin >> n >> m;
  vector<vector<long long> > dpG(n + 1, vector<long long>(m + 1, 0));
  vector<vector<long long> > dpD(n + 1, vector<long long>(m + 1, 0));
  for (int i = 2; i <= m; ++i) {
    dpG[1][i] = 1;
    dpD[1][i] = 1;
  }
  for (int i = 2; i <= n; ++i) {
    vector<long long> pref1G(m + 1, 0), pref2G(m + 1, 0);
    for (long long j = 2; j <= m; ++j) {
      pref1G[j] = (pref1G[j - 1] + dpG[i - 1][j]) % MOD;
      pref2G[j] = (pref2G[j - 1] + dpG[i - 1][j] * j) % MOD;
    }
    for (long long j = 2; j <= m; ++j) {
      dpG[i][j] = (pref1G[j] * (j + 1)) % MOD;
      dpG[i][j] -= pref2G[j];
      if (dpG[i][j] < 0) dpG[i][j] += MOD;
      dpD[i][j] = (pref1G[j - 1] * (j + 1)) % MOD;
      dpD[i][j] -= pref2G[j - 1];
      if (dpD[i][j] < 0) dpD[i][j] += MOD;
    }
  }
  vector<vector<long long> > prefG(m + 1, vector<long long>(n + 1, 0));
  vector<vector<long long> > prefD(m + 1, vector<long long>(n + 1, 0));
  for (int i = 2; i <= m; ++i) {
    for (int j = 1; j <= n; ++j) {
      prefG[i][j] = (prefG[i][j - 1] + dpG[j][i]) % MOD;
      prefD[i][j] = (prefD[i][j - 1] + dpD[j][i]) % MOD;
    }
  }
  long long odp = 0;
  for (int i = 1; i <= n; ++i) {
    for (long long j = 2; j <= m; ++j) {
      long long t = prefG[j][i] * prefD[j][n - i + 1] % MOD * (m - j + 1) % MOD;
      odp = (odp + t) % MOD;
    }
  }
  cout << odp;
  return 0;
}
