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 V = 2100;
const int N = 2000;
const int P = 1000000007;
int dp[V][V], s[V], sn[V], ls[V][V];
void init() {
  for (int i = 0; i <= N; i++) dp[1][i] = 1;
  s[0] = dp[1][0];
  sn[0] = (long long)dp[1][0] * (N - 0 + 1) % P;
  for (int j = 1; j <= N; j++) {
    s[j] = s[j - 1] + dp[1][j];
    s[j] %= P;
    sn[j] = sn[j - 1] + (long long)dp[1][j] * (N - j + 1) % P;
    sn[j] %= P;
  }
  for (int i = 2; i <= N; i++) {
    for (int j = 0; j <= N; j++) {
      dp[i][j] = sn[j] - (long long)(N - j) * s[j] % P;
      dp[i][j] = (dp[i][j] % P + P) % P;
    }
    s[0] = dp[i][0];
    sn[0] = (long long)dp[i][0] * (N - 0 + 1) % P;
    for (int j = 1; j <= N; j++) {
      s[j] = s[j - 1] + dp[i][j];
      s[j] %= P;
      sn[j] = sn[j - 1] + (long long)dp[i][j] * (N - j + 1) % P;
      sn[j] %= P;
    }
  }
  memset(ls, 0, sizeof(ls));
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= N; j++) ls[i][j] = (ls[i - 1][j] + dp[i][j]) % P;
}
int n, m;
int main() {
  init();
  while (~scanf("%d%d", &n, &m)) {
    if (m < 2) {
      puts("0");
      continue;
    }
    int ret = 0;
    for (int i = 0; i < n; i++) {
      s[0] = ls[n - i - 1][0];
      sn[0] = (long long)ls[n - i - 1][0] * (N - 0 + 1) % P;
      for (int j = 1; j <= m; j++) {
        s[j] = (s[j - 1] + ls[n - i - 1][j]) % P;
        sn[j] = sn[j - 1] + (long long)ls[n - i - 1][j] * (N - j + 1) % P;
        sn[j] %= P;
      }
      for (int j = 0; j <= m - 2; j++) {
        int way = m - 2 - j + 1;
        way = (long long)way * ls[i + 1][j] % P;
        if (j != 0) {
          way = (long long)way *
                (sn[j - 1] + 1 - (long long)(N - j) * s[j - 1]) % P;
          way = (way % P + P) % P;
        }
        ret += way;
        ret %= P;
      }
    }
    printf("%d\n", ret);
  }
}
