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 MX = 2005;
const int MOD = 1000000007;
long long D[MX][MX], T[MX][MX], E[MX][MX];
long long tot[MX];
int main() {
  int N, M;
  scanf("%d%d", &N, &M);
  for (int j = 2; j <= M; j++) D[1][j] = 1;
  for (int i = 2; i <= N; i++) {
    for (int j = 2; j <= M; j++) D[i][j] = (D[i][j - 1] + D[i - 1][j]) % MOD;
    for (int j = 2; j <= M; j++) D[i][j] = (D[i][j] + D[i][j - 1]) % MOD;
  }
  for (int j = 2; j <= M; j++) {
    for (int i = 1; i <= N; i++) D[i][j] = (D[i][j] + D[i - 1][j]) % MOD;
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= M; j++) E[i][j] = (E[i][j - 1] + D[i][j]) % MOD;
    for (int j = 1; j <= M; j++) E[i][j] = (E[i][j - 1] + E[i][j]) % MOD;
    for (int j = 1; j <= M; j++) E[i][j] = (E[i][j] - D[i][j] + MOD + 1) % MOD;
  }
  long long ans = 0;
  for (int j = 2; j <= M; j++) {
    long long tmp = D[N][j];
    for (int k = 1; k < N; k++) {
      tmp = (tmp + D[k][j] * E[N - k][j]) % MOD;
    }
    ans = (ans + tmp * (M - j + 1)) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}
