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 INF = 2147483647;
const long long LLINF = 9223372036854775807LL;
const int mod = 1000000007;
struct num {
  int x;
  num() : x(0) {}
  num(int x) : x(x) {}
  num operator+(const num& o) const {
    num res(x + o.x);
    if (res.x >= mod) res.x -= mod;
    return res;
  }
  num operator-(const num& o) const {
    num res(x - o.x);
    if (res.x < 0) res.x += mod;
    return res;
  }
  num operator*(const num& o) const { return num(x * (long long)o.x % mod); }
};
const int maxn = 2010;
num f[maxn][maxn];
num ff[maxn][maxn];
num fsum[maxn][maxn];
int main() {
  for (int j = 2; j < maxn; ++j) {
    f[1][j] = num(1);
    ff[1][j] = f[1][j] * num(j) + ff[1][j - 1];
    fsum[1][j] = f[1][j] + fsum[1][j - 1];
  }
  for (int i = 2; i < maxn; ++i) {
    for (int j = 2; j < maxn; ++j) {
      f[i][j] = num(1) + num(j + 1) * fsum[i - 1][j] - ff[i - 1][j];
      ff[i][j] = f[i][j] * num(j) + ff[i][j - 1];
      fsum[i][j] = fsum[i][j - 1] + f[i][j];
    }
  }
  int n, m;
  scanf("%d%d", &n, &m);
  num res(0);
  for (int i = 1; i <= n; ++i) {
    num curcnt(0);
    for (int len = 1; len <= m; ++len) {
      num curadd =
          num(m - len + 1) * f[i][len] *
          (fsum[n - i][len - 1] * num(len + 1) - ff[n - i][len - 1] + num(1));
      curcnt = curcnt + curadd;
    }
    res = res + curcnt;
  }
  printf("%d\n", res.x);
  return 0;
}
