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 P = 1000000000 + 7;
long long fact[10001], tcaf[10001];
long long delta[2001][2001];
long long well[2001][2001], cyl[2001];
long long deg(long long x, int d) {
  if (d == 0) {
    return 1;
  }
  long long y = deg(x, d / 2);
  y *= y;
  y %= P;
  if (d % 2) {
    y *= x;
    y %= P;
  }
  return y;
}
long long inv(long long x) { return deg(x, P - 2); }
long long ways(int x, int y) {
  if (y == 1) {
    return 1;
  } else {
    return fact[x + 2 * y - 5] * tcaf[2 * y - 3] % P * tcaf[x - 2] % P;
  }
}
int main() {
  int N, M;
  cin >> N >> M;
  fact[0] = tcaf[0] = 1;
  for (int i = 1; i <= 10000; ++i) {
    fact[i] = (fact[i - 1] * i) % P;
    tcaf[i] = inv(fact[i]);
  }
  for (int x = 2; x <= M; ++x) {
    well[x][1] = 1;
    for (int y = 2; y <= N; ++y) {
      delta[x][y] = delta[x - 1][y] + ways(x, y);
      delta[x][y] %= P;
      well[x][y] = well[x][y - 1] + delta[x][y];
      well[x][y] %= P;
    }
  }
  for (int x = 2; x <= M; ++x) {
    for (int y = 1; y <= N; ++y) {
      cyl[x] += well[x][y] * (well[x][N - y + 1] - well[x][N - y] + P) % P;
      cyl[x] %= P;
    }
  }
  long long ans = 0;
  for (int x = 2; x <= M; ++x) {
    ans += cyl[x] * (M - x + 1);
    ans %= P;
  }
  cout << ans << '\n';
  return 0;
}
