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;
char _;
const int maxN = 2000 + 15;
const int maxM = 100 * 100 + 5;
const int mod = 1000 * 1000 * 1000 + 7;
const int base = 701;
const int SQ = 500;
const int maxL = 17;
int dp[maxN][maxN];
int sum[maxN][maxN];
int n, m;
void upd(int i) {
  int res = 0;
  for (int j = 1; j < m + 1; j++) {
    res = (res + dp[i][j]) % mod;
    sum[i][j] = (sum[i][j - 1] + res) % mod;
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i < m + 1; i++) sum[0][i] = 1;
  for (int i = 2; i < m + 1; i++) dp[1][i] = 1;
  upd(1);
  for (int i = 2; i < n + 1; i++) {
    for (int j = 1; j < m + 1; j++) dp[i][j] = sum[i - 1][j];
    upd(i);
  }
  int ans = 0;
  for (int i = 1; i < m + 1; i++) {
    int ted = (m - i + 1);
    int par = 0, res = 0;
    for (int j = n; j > 0; j--) {
      int tmp = (sum[n - j][i] + mod - dp[n - j][i]) % mod;
      par = (par + tmp) % mod;
      res = (res + par) % mod;
      ans = (ans + (1LL) * dp[j][i] * ((1LL) * res * ted % mod) % mod) % mod;
    }
  }
  cout << ans << endl;
  return 0;
}
