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 = 1000000007;
int add(int a, int b) { return a + b >= MOD ? a + b - MOD : a + b; }
int mul(int a, int b) { return (long long)a * b % MOD; }
int sub(int a, int b) { return a < b ? a - b + MOD : a - b; }
const int MAXN = 2001;
int n, m, expand[2][MAXN], shrink[2][MAXN];
int sumei[MAXN], sumiei[MAXN];
int sumEi[MAXN], sumiEi[MAXN];
int sumsi[MAXN], sumisi[MAXN];
int main() {
  scanf("%d%d", &n, &m);
  int t = 0, res = 0;
  for (int width = (2); width < (m + 1); width++) {
    expand[t][width] = 1;
    res = add(res, mul(expand[t][width], (m - width + 1) * n));
  }
  for (int height = (2); height < (n + 1); height++) {
    for (int width = (2); width < (m + 1); width++) {
      sumei[width] = add(sumei[width - 1], expand[t][width]);
      sumiei[width] = add(sumiei[width - 1], mul(width, expand[t][width]));
      sumEi[width] =
          add(sumEi[width - 1], mul(expand[t][width], m - width + 1));
      sumiEi[width] = add(sumiEi[width - 1],
                          mul(expand[t][width], width * (m - width + 1)));
      sumsi[width] = add(sumsi[width - 1], shrink[t][width]);
      sumisi[width] = add(sumisi[width - 1], mul(width, shrink[t][width]));
    }
    t ^= 1;
    memset(expand[t], 0, sizeof(int) * (m + 1));
    memset(shrink[t], 0, sizeof(int) * (m + 1));
    for (int width = (2); width < (m + 1); width++) {
      expand[t][width] = sub(mul(sumei[width], width + 1), sumiei[width]);
      res = add(res, mul(expand[t][width], (m - width + 1) * (n - height + 1)));
      shrink[t][width] =
          add(sub(sub(sumiEi[m], sumiEi[width]),
                  mul(sub(sumEi[m], sumEi[width]), width - 1)),
              sub(sub(sumisi[m], sumisi[width - 1]),
                  mul(sub(sumsi[m], sumsi[width - 1]), width - 1)));
      res = add(res, mul(shrink[t][width], n - height + 1));
    }
  }
  printf("%d\n", res);
  return 0;
}
