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 MAX = 2100;
const long long INF = 1.0e9;
const double EPS = 1.0e-8;
const long long MOD = 1000000007;
const double PI = acos(-1.0);
long long sum[MAX][MAX] = {0};
long long dp[MAX][MAX] = {0};
int dblcmp(double x) {
  if (fabs(x) < EPS) return 0;
  return x < 0 ? -1 : 1;
}
int main() {
  int i, j;
  long long tmp;
  for (i = 2; i < MAX; i++) dp[1][i] = sum[1][i] = 1;
  for (i = 2; i < MAX; i++) {
    tmp = 0;
    for (j = 2; j < MAX; j++) {
      dp[i][j] = dp[i][j - 1];
      tmp += dp[i - 1][j];
      if (tmp >= MOD) tmp -= MOD;
      dp[i][j] += tmp;
      if (dp[i][j] >= MOD) dp[i][j] -= MOD;
    }
  }
  for (i = 2; i < MAX; i++) {
    tmp = 0;
    for (j = 2; j < MAX; j++) {
      sum[i][j] = dp[i][j - 1];
      tmp += dp[i - 1][j];
      if (tmp >= MOD) tmp -= MOD;
      sum[i][j] += tmp - dp[i - 1][j];
      if (sum[i][j] >= MOD) sum[i][j] -= MOD;
      if (sum[i][j] < 0) sum[i][j] += MOD;
    }
  }
  for (i = 1; i < MAX; i++) {
    for (j = 0; j < MAX; j++) {
      dp[i][j] += dp[i - 1][j];
      if (dp[i][j] >= MOD) dp[i][j] -= MOD;
      sum[i][j] += sum[i - 1][j];
      if (sum[i][j] >= MOD) sum[i][j] -= MOD;
    }
  }
  int n, m;
  while (scanf("%d%d", &n, &m) != EOF) {
    long long ans = 0;
    for (i = 1; i <= n; i++) {
      for (j = 2; j <= m; j++) {
        tmp = dp[i][j] * sum[n - i + 1][j] % MOD * (m - j + 1);
        ans = (ans + tmp) % MOD;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
