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;
const int N = 2005;
int n, m;
long long f[N][N], g[N][N], sum[N][N], sumu[N][N], sumf[N][N], sumk[N][N];
int main() {
  scanf("%d %d", &n, &m);
  memset(f, 0, sizeof(f));
  memset(sum, 0, sizeof(sum));
  memset(sumu, 0, sizeof(sumu));
  for (int j = 1; j <= n; ++j) {
    for (int i = 2; i <= m; ++i) {
      if (j == 1) {
        f[i][j] = 1;
      } else {
        f[i][j] =
            (((i + 1) * sum[i][j - 1] % MOD - sumu[i][j - 1]) % MOD + MOD) %
            MOD;
      }
      sum[i][j] = (sum[i - 1][j] + f[i][j]) % MOD;
      sumu[i][j] = (sumu[i - 1][j] + i * f[i][j] % MOD) % MOD;
    }
  }
  memset(g, 0, sizeof(g));
  memset(sum, 0, sizeof(sum));
  memset(sumu, 0, sizeof(sumu));
  for (int j = 1; j <= n; ++j) {
    for (int i = 2; i <= m; ++i) {
      if (j == 1) {
        g[i][j] = 1;
      } else {
        g[i][j] =
            (((i + 1) * sum[i - 1][j - 1] % MOD - sumu[i - 1][j - 1]) % MOD +
             MOD) %
            MOD;
      }
      sum[i][j] = (sum[i - 1][j] + f[i][j]) % MOD;
      sumu[i][j] = (sumu[i - 1][j] + i * f[i][j] % MOD) % MOD;
    }
  }
  memset(sumf, 0, sizeof(sumf));
  memset(sumk, 0, sizeof(sumk));
  for (int i = 2; i <= m; ++i) {
    for (int j = 1; j <= n; ++j) {
      sumf[i][j] = (sumf[i][j - 1] + f[i][j]) % MOD;
      sumk[i][j] = (sumk[i][j - 1] + j * f[i][j] % MOD) % MOD;
    }
  }
  long long ans = 0;
  for (int v = 2; v <= m; ++v) {
    for (int i = 1; i <= n; ++i) {
      ans += (m - v + 1) * g[v][i] % MOD *
             ((((n - i + 2) * sumf[v][n - i + 1] % MOD - sumk[v][n - i + 1]) %
                   MOD +
               MOD) %
              MOD) %
             MOD;
      ans %= MOD;
    }
  }
  cout << ans << endl;
  return 0;
}
