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;
long long dp[2005][2005], sum[2005][2005];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int j = 2; j <= m; j++) dp[j][1] = 1;
  for (int i = 2; i <= n; i++) {
    long long sum1 = 0, sum2 = 0;
    for (int j = 2; j <= m; j++) {
      sum1 = (sum1 + dp[j][i - 1]) % mod;
      sum2 = (sum2 + sum1) % mod;
      dp[j][i] = sum2;
    }
  }
  for (int j = 2; j <= m; j++) {
    for (int i = 1; i <= n; i++) {
      sum[j][i] = (sum[j][i - 1] + dp[j][i]) % mod;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long tot = 0, tmp = 0;
    for (int j = 2; j <= m; j++) {
      long long cnt = m - j + 1;
      long long res = 0;
      int mxr = n - i;
      if (i == 1) {
        res = (res + tot * cnt) % mod;
        res = (res + cnt) % mod;
      } else if (i == n) {
        res = (res + sum[j][i] * cnt) % mod;
      } else {
        res = (res + sum[j][i] * tot % mod * cnt) % mod;
        res = (res + sum[j][i] * cnt) % mod;
      }
      ans = (ans + res) % mod;
      tmp = (tmp + sum[j][mxr]) % mod;
      tot = (tot + sum[j][mxr] + tmp) % mod;
    }
  }
  printf("%lld\n", ans);
}
