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 long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
int n, m, ans;
int dp[2010][2010], pd[2010][2010];
int main() {
  scanf("%d%d", &n, &m);
  for (int j = 0; j < m + 1; j++) dp[0][j] = 1, pd[0][j] = 1;
  for (int i = 1; i < n + 1; i++) {
    int sm = 1;
    dp[i][0] = 1;
    pd[i][0] = 0;
    for (int j = 1; j < m + 1; j++) {
      sm = (sm + dp[i - 1][j]) % mod;
      dp[i][j] = (dp[i][j - 1] + sm) % mod;
      pd[i][j] = (dp[i][j] - dp[i - 1][j]) % mod;
    }
  }
  for (int j = 0; j < m + 1; j++)
    for (int i = 1; i < n + 1; i++) {
      dp[i][j] = (dp[i - 1][j] + dp[i][j]) % mod;
      pd[i][j] = (pd[i - 1][j] + pd[i][j]) % mod;
    }
  for (int c = 0; c < m; c++)
    for (int k = 1; k < n + 1; k++)
      ans = (ans + 1ll * dp[n - k][c] * pd[k - 1][c] % mod * (m - c - 1)) % mod;
  if (ans < 0) ans += mod;
  printf("%d\n", ans);
}
