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 = 1000000007;
inline int add(int a, int b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline int mul(int a, int b) { return (a * (long long)b) % mod; }
inline int sub(int a, int b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
int dp[2010][2010];
int dp2[2010][2010];
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  for (int j = (2); j <= (m); j++) dp[1][j] = 1;
  for (int i = (2); i <= (n); i++) {
    int s = 0, sj = 0;
    for (int j = (1); j <= (m); j++) {
      s = add(s, dp[i - 1][j]);
      sj = add(sj, mul(dp[i - 1][j], j));
      dp[i][j] = sub(mul(j + 1, s), sj);
    }
  }
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (m); j++) {
      dp[i][j] = mul(dp[i][j], m - j + 1);
    }
  for (int i = (1); i <= (n); i++) {
    int s = 0, sj = 0;
    int ss = 0, ssj = 0;
    for (int j = (m); j >= (2); j--) {
      dp2[i][j] = sub(sj, mul(j - 1, s));
      s = add(s, dp[i - 1][j]);
      sj = add(sj, mul(dp[i - 1][j], j));
      ss = add(ss, dp2[i - 1][j]);
      ssj = add(ssj, mul(dp2[i - 1][j], j));
      dp2[i][j] = add(dp2[i][j], sub(ssj, mul(j - 1, ss)));
    }
  }
  int ans = 0;
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (m); j++) {
      ans = add(ans, mul(n - i + 1, dp[i][j]));
      ans = add(ans, mul(n - i + 1, dp2[i][j]));
    }
  printf("%d\n", ans);
}
