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 mod = 1e9 + 7;
int exp(int x, int n) {
  int ret = 1;
  while (n) {
    if (n & 1) ret = 1LL * ret * x % mod;
    x = 1LL * x * x % mod;
    n >>= 1;
  }
  return ret;
}
int inv(int x) { return exp(x, mod - 2); }
int fact[10010], invf[10010];
int comb(int n, int k) {
  return 1LL * fact[n] * invf[k] % mod * invf[n - k] % mod;
}
int N, M;
int prec[2010][2010];
int main() {
  fact[0] = 1;
  for (int i = 1; i < 10010; i++) {
    fact[i] = 1LL * fact[i - 1] * i % mod;
  }
  for (int i = 0; i < 10010; i++) {
    invf[i] = inv(fact[i]);
  }
  scanf("%d %d", &N, &M);
  for (int i = 2; i <= M; i++) {
    prec[i][0] = 1;
    for (int j = 1; j < N; j++) {
      prec[i][j] = comb(i + 2 * j - 2, 2 * j);
    }
    for (int j = 1; j < N; j++) {
      prec[i][j] += prec[i][j - 1];
      prec[i][j] %= mod;
    }
  }
  int ans = 0;
  for (int i = 0; i < N; i++) {
    for (int j = 2; j <= M; j++) {
      ans += 1LL * (M - j + 1) * prec[j][i] % mod *
             (prec[j][N - 1 - i] - (i == N - 1 ? 0 : prec[j][N - 2 - i])) % mod;
      ans %= mod;
    }
  }
  printf("%d", ans);
}
