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 mod = 1e9 + 7;
int add(int x, int y) {
  x += y;
  if (x >= mod) return x - mod;
  return x;
}
int sub(int x, int y) {
  x -= y;
  if (x < 0) return x + mod;
  return x;
}
int mul(int x, int y) { return (long long)x * y % mod; }
const int MAXN = 2020;
int n, m;
int dp[MAXN][MAXN], dpm[MAXN][MAXN], rast[MAXN], strt[MAXN];
void postavi(int red) {
  int k = m - 1;
  for (int i = (0); i < (m - 1); ++i) {
    rast[i] = mul(dp[red][i], k);
    strt[i] = dp[red][i];
    if (i) strt[i] = add(strt[i], strt[i - 1]);
    if (i) rast[i] = add(rast[i], rast[i - 1]);
    k--;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = (0); i < (m - 1); ++i) {
    dp[0][i] = 1;
    dpm[0][i] = 1;
  }
  postavi(0);
  for (int i = (1); i < (n); ++i) {
    for (int j = (0); j < (m - 1); ++j) {
      dp[i][j] = 1;
      dp[i][j] = add(rast[j], dp[i][j]);
      dp[i][j] = sub(dp[i][j], mul(strt[j], m - 2 - j));
      dpm[i][j] = 1;
      if (j) {
        dpm[i][j] = add(rast[j - 1], dpm[i][j]);
        dpm[i][j] = sub(dpm[i][j], mul(strt[j - 1], m - 2 - j));
      }
    }
    postavi(i);
  }
  int sol = 0;
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m - 1); ++j) {
      sol = add(sol, mul(m - 1 - j, mul(dpm[i][j], dp[n - i - 1][j])));
    }
  }
  cout << sol;
}
