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;
int n, m;
int C[6001][6001];
long long mod = 1000000007LL;
long long A[2001][2001];
long long B[2001][2001];
long long u[2001], su[2001];
int MAIN() {
  memset(C, 0, sizeof(C));
  for (int i = 0; i <= 6000; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
  }
  int n, m;
  while (cin >> n >> m) {
    memset(A, 0, sizeof(A));
    memset(B, 0, sizeof(B));
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        A[i][j] = C[j + 2 * i - 2][2 * i];
        A[i][j] += A[i - 1][j];
        if (i == 1) A[i][j]++;
        A[i][j] %= mod;
      }
    }
    for (int i = 1; i <= n; i++)
      for (int j = 2; j <= m; j++) {
        B[i][j] = A[i][j] - 2 * A[i][j - 1] + A[i][j - 2];
        if (j == 2) B[i][j]++;
        B[i][j] %= mod;
        B[i][j] += mod;
        B[i][j] %= mod;
      }
    long long ans = 0;
    for (int len = 2; len <= m; len++) {
      long long t = 0;
      for (int i = 0; i <= n; i++) {
        if (i == 0)
          u[i] = 1;
        else
          u[i] = (A[i][len] - B[i][len] + mod) % mod;
        su[i] = u[i];
        if (i > 0) su[i] += su[i - 1];
        su[i] %= mod;
      }
      for (int i = 0; i <= n - 1; i++) {
        t += u[i] * su[n - 1 - i];
        t %= mod;
      }
      ans += t * (m - len + 1);
      ans %= mod;
    }
    cout << ans << endl;
  }
  return 0;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  return MAIN();
}
