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 mod = 1000000007;
long long fact[2005 + 2005], inv[2005 + 2005];
long long dp[2005][2005], dwn[2005], up[2005];
int N, M;
long long bigmod(long long x, long long pw) {
  long long ret = 1;
  while (pw > 0) {
    if (pw & 1) ret = ret * x % mod;
    x = x * x % mod;
    pw >>= 1;
  }
  return ret;
}
long long nCr(int n, int r) {
  if (r > n) return 0;
  return fact[n] * (inv[n - r] * inv[r] % mod) % mod;
}
int main() {
  fact[0] = inv[0] = 1;
  for (int i = 1; i < 2005 + 2005; i++) {
    fact[i] = i * fact[i - 1] % mod;
    inv[i] = bigmod(fact[i], mod - 2);
  }
  for (int i = 0; i < 2005; i++)
    for (int j = 0; j < 2005; j++) dp[j][i] = nCr(j + 2 * i, 2 * i);
  scanf("%d", &N);
  scanf("%d", &M);
  long long ans = 0;
  for (int i = 0; i < M; i++) {
    up[i] = dp[i][0];
    for (int j = 0; j < N; j++) dwn[i] = (dwn[i] + dp[i][j]) % mod;
  }
  for (int r = 0; r < N; r++) {
    for (int c = 0; c + 2 <= M; c++)
      ans = (ans + (up[c] * dwn[c] % mod) * (M - c - 1)) % mod;
    for (int i = 0; i < M; i++) {
      up[i] = (up[i] + dp[i][r + 1] - dp[i][r]) % mod;
      dwn[i] = dwn[i] - dp[i][N - r - 1];
      if (up[i] < 0) up[i] += mod;
      if (dwn[i] < 0) dwn[i] += mod;
    }
  }
  cout << ans << endl;
}
