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 double eps = 1e-8;
const double pi = acos(-1.0);
int dblcmp(double d) {
  if (fabs(d) < eps) return 0;
  return d > eps ? 1 : -1;
}
const int mod = 1000000007;
int n, m;
long long dp[2111][2111][2];
long long sum[2111], sm[2111];
int main() {
  int i, j, k, cas;
  cin >> n >> m;
  memset((dp), 0, sizeof(dp));
  dp[0][0][0] = 1;
  for (i = 1; i <= n; i++) {
    dp[i][0][0] += dp[i - 1][0][0];
    memset((sum), 0, sizeof(sum));
    memset((sm), 0, sizeof(sm));
    for (j = 2; j <= m; j++) dp[i][j][0] += dp[i - 1][0][0];
    for (j = 0; j <= m; j++) {
      sum[j] = (dp[i - 1][j][0] + dp[i - 1][j][1]) * j * (-1LL);
      sum[j] %= mod;
      sum[j] += mod;
      sum[j] %= mod;
      sm[j] = (dp[i - 1][j][0] + dp[i - 1][j][1]);
      sm[j] %= mod;
    }
    for (j = 1; j <= m; j++) {
      sum[j] += sum[j - 1];
      sum[j] %= mod;
      sm[j] += sm[j - 1];
      sm[j] %= mod;
    }
    for (j = 2; j <= m; j++) {
      dp[i][j][0] += (sum[j - 1] - sum[1]);
      dp[i][j][0] %= mod;
      dp[i][j][0] += mod;
      dp[i][j][0] %= mod;
      dp[i][j][0] += (sm[j - 1] - sm[1]) * (j + 1LL);
      dp[i][j][0] %= mod;
      dp[i][j][0] += mod;
      dp[i][j][0] %= mod;
      dp[i][j][1] += dp[i - 1][j][0];
      dp[i][j][1] += dp[i - 1][j][1];
      dp[i][j][1] %= mod;
    }
  }
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    memset((sum), 0, sizeof(sum));
    memset((sm), 0, sizeof(sm));
    for (j = 0; j <= m; j++) {
      sum[j] = (dp[n - i][j][0] + dp[n - i][j][1]) * j * (-1LL);
      sum[j] %= mod;
      sum[j] += mod;
      sum[j] %= mod;
      sm[j] = (dp[n - i][j][0] + dp[n - i][j][1]);
      sm[j] %= mod;
    }
    for (j = 1; j <= m; j++) {
      sum[j] += sum[j - 1];
      sum[j] %= mod;
      sm[j] += sm[j - 1];
      sm[j] %= mod;
    }
    for (j = 2; j <= m; j++) {
      long long tmp = 1;
      tmp += (sum[j] - sum[1]);
      tmp %= mod;
      tmp += (sm[j] - sm[1]) * (j + 1LL);
      tmp %= mod;
      tmp += mod;
      tmp %= mod;
      ans += (((dp[i][j][0] * tmp) % mod) * (m - j + 1LL)) % mod;
      ans %= mod;
    }
  }
  cout << ans << endl;
  return 0;
}
