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 N = 2222;
const int MOD = int(1e9) + 7;
int n, m;
int f[N][N];
int g[N][N];
int h[N][N];
int comb[N][N];
inline void add(int &x, int t) {
  if ((x += t) >= MOD) {
    x -= MOD;
  }
}
int main() {
  for (int i = 0; i < N; i++) {
    comb[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      comb[i][j] = comb[i - 1][j - 1] + comb[i - 1][j];
      if (comb[i][j] >= MOD) {
        comb[i][j] -= MOD;
      }
    }
  }
  scanf("%d%d", &n, &m);
  for (int j = 2; j <= m; j++) {
    g[1][j] = f[1][j] = 1;
    add(g[1][j], g[1][j - 1]);
  }
  for (int i = 2; i <= n; i++) {
    int now = 0;
    for (int j = 1; j <= m; j++) {
      add(now, g[i - 1][j]);
      g[i][j] = f[i][j] = now;
      add(g[i][j], g[i][j - 1]);
    }
  }
  for (int j = 1; j <= m; j++) {
    for (int i = 1; i <= n; i++) {
      h[i][j] = f[i][j];
      add(h[i][j], h[i - 1][j]);
    }
  }
  memcpy(g, h, sizeof(g));
  memset(h, 0, sizeof(h));
  for (int i = 1; i <= n; i++) {
    int now = 0, tmp = 0;
    for (int j = 1; j <= m; j++) {
      add(now, tmp);
      h[i][j] = now;
      add(now, g[i][j]);
      add(tmp, g[i][j]);
    }
  }
  int result = 0;
  for (int row = 1; row <= n; row++) {
    for (int length = 1; length <= m; length++) {
      int now = (h[row - 1][length] + 1) % MOD;
      now = 1LL * now * g[n - row + 1][length] % MOD;
      now = 1LL * now * (m - length + 1) % MOD;
      add(result, now);
    }
  }
  cout << result << endl;
}
