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>
#pragma comment(linker, "/stack:200000000")
using namespace std;
const int mod = 1000000007;
const int maxn = 100010;
const int inf = 2e9;
const double eps = 1e-7;
inline int gcd(int a, int b) {
  if (a < b) swap(a, b);
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
int dp[2002][2002];
int dpp[2002][2002];
int pref[2002][2002];
int mul(int a, int b) { return ((long long)a * (long long)b) % mod; }
int main() {
  int n, m;
  scanf("%d %d", &(n), &(m));
  ;
  for (int i = (2); i < (m + 1); i++) {
    dp[0][i] = dpp[0][i] = 1;
    pref[0][i] = pref[0][i - 1] + dp[0][i];
  }
  for (int i = (1); i < (n); i++) {
    int cur_sum = 0;
    for (int j = (2); j < (m + 1); j++) {
      dp[i][j] += (dp[i][j - 1] + pref[i - 1][j] + (j == 2 ? 1 : 0)) % mod;
      pref[i][j] = (pref[i][j - 1] + dp[i][j]) % mod;
      dpp[i][j] = dp[i][j] - dp[i - 1][j];
    }
  }
  int ans = 0;
  for (int i = (0); i < (n); i++)
    for (int j = (2); j < (m + 1); j++) {
      ans += mul(mul(dp[i][j], dpp[n - 1 - i][j]), m - j + 1), ans %= mod;
    }
  cout << ans << endl;
}
