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 MN = 2005, inf = 1000000005, mod = 1000000007;
const long long INF = 1000000000000000005LL;
long long dp[MN][MN], pref[MN][MN];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  if (m < 2) {
    printf("0");
    return 0;
  }
  for (int k = 1; k <= n; ++k) {
    long long suma = 0LL, wsuma = 0LL;
    for (int l = 2; l <= m; ++l) {
      suma += dp[k - 1][l];
      suma %= mod;
      wsuma += suma;
      wsuma %= mod;
      dp[k][l] = (wsuma + 1LL + mod) % mod;
    }
  }
  for (int k = n; k > 1; --k)
    for (int l = 2; l <= m; ++l) dp[k][l] -= dp[k - 1][l];
  long long ans = 0LL;
  for (int i = 1; i <= n; ++i)
    for (int l = 2; l <= m; ++l) pref[i][l] = (pref[i - 1][l] + dp[i][l]) % mod;
  for (int i = 1; i <= n; ++i)
    for (int l = 2; l <= m; ++l) {
      ans +=
          ((((dp[i][l] * pref[n - i + 1][l]) % mod) * 1LL * (m - l + 1)) % mod);
      ans %= mod;
    }
  printf("%lld", ans);
}
