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 maxn = 2 * 1000 + 3;
const long long MOD = 1000000007;
long long dp[maxn][maxn], sum[maxn][maxn];
long long memo(int i, int j) {
  if (i == 0) {
    dp[i][j] = 1ll;
    return 1ll;
  }
  if (j < 2) {
    dp[i][j] = 0ll;
    return 0ll;
  }
  if (i == 1) {
    dp[i][j] = (j * (j - 1) / 2) % MOD;
    return (j * (j - 1) / 2) % MOD;
  }
  if (dp[i][j] == -1)
    dp[i][j] =
        (memo(i - 1, j) + ((2 * memo(i, j - 1)) % MOD) - memo(i, j - 2)) % MOD;
  return dp[i][j];
}
int main() {
  memset(dp, -1, sizeof dp);
  int n, m;
  cin >> n >> m;
  memo(n, m);
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= m; j++)
      if (dp[i][j] == -1) dp[i][j]++;
  for (int i = 0; i <= m; i++) sum[0][i] = 1ll;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= m; j++)
      sum[i][j] = (((sum[i][j] + sum[i - 1][j]) % MOD) + dp[i][j]) % MOD;
  long long ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 2; j <= m; j++)
      ans =
          (ans + ((sum[i - 1][j] *
                   (((2 * sum[n - i][j - 1] - sum[n - i][j - 2]) + MOD) % MOD) %
                   MOD) *
                  (m - j + 1)) %
                     MOD) %
          MOD;
  cout << (ans + MOD) % MOD << endl;
  return 0;
}
