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;
long long cumdp[2005];
long long cumpd[2005];
long long dp[2005][2005];
long long pd[2005][2005];
const long long mod = 1e9 + 7;
int main() {
  long long n, m;
  cin >> n >> m;
  for (int width = 2; width <= m; width++) {
    dp[1][width] = 1;
    pd[1][width] = 1;
  }
  for (int len = 2; len <= n; len++) {
    long long sum = 0;
    for (int width = 2; width <= m; width++) {
      sum += dp[len - 1][width];
      sum %= mod;
      dp[len][width] = dp[len][width - 1] + sum;
      pd[len][width] = dp[len][width] - dp[len - 1][width];
    }
  }
  long long ans = 0;
  for (int base = 1; base <= m; base++) {
    for (int len = 1; len <= n; len++) {
      cumdp[len] = cumdp[len - 1] + dp[len][base];
      cumdp[len] %= mod;
      cumpd[len] = cumpd[len - 1] + pd[len][base];
      cumpd[len] %= mod;
    }
    long long cur = 0;
    for (int cen = 1; cen <= n; cen++) {
      cur += (cumdp[cen] * cumpd[n - cen + 1]) % mod;
      cur %= mod;
    }
    cur *= (m - base + 1);
    cur %= mod;
    ans += cur;
    ans %= mod;
  }
  cout << ans << "\n";
}
