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>
#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char ch = getchar();
  long long f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f *= -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  x *= f;
}
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int maxn = 2e3 + 7;
const int maxm = 25;
const int mod = 1e9 + 7;
long long dp[maxn][maxn];
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) dp[1][i] = 1;
  for (int i = 2; i <= n; ++i) {
    long long sum = 0;
    dp[i][1] = 1;
    for (int j = 2; j <= m; ++j) {
      sum = (sum + dp[i - 1][j]) % mod;
      dp[i][j] = (dp[i][j - 1] + sum) % mod;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 2; j <= m; ++j)
      ans = (ans + ((dp[i][j] - dp[i - 1][j]) * dp[n - i + 1][j] % mod) *
                       (m - j + 1)) %
            mod;
  cout << ans << '\n';
}
