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;
void Get(int &T) {
  char C;
  bool F = 0;
  for (; C = getchar(), C < '0' || C > '9';)
    if (C == '-') F = 1;
  for (T = C - '0'; C = getchar(), C >= '0' && C <= '9'; T = T * 10 + C - '0')
    ;
  F && (T = -T);
}
long long F[2005][2005];
long long G[2005][2005];
int N, M;
void Init() {
  Get(N);
  Get(M);
}
void Debug() {
  for (int i = 1; i <= N; i++) {
    for (int k = 1; k <= M; k++) printf("%lld ", F[i][k]);
    puts("");
  }
  puts("");
  puts("");
  for (int i = 1; i <= N; i++) {
    for (int k = 1; k <= M; k++) printf("%lld ", G[i][k]);
    puts("");
  }
}
long long Ans;
void Work() {
  F[1][2] = 1;
  for (int i = 1; i <= N; i++)
    for (int k = 1; k <= M; k++) {
      (F[i][k] += F[i][k - 1]) %= (1000000007);
      (F[i + 1][k] = F[i + 1][k - 1] + F[i][k]) %= (1000000007);
      (G[i][k] = F[i][k] - F[i - 1][k]) %= (1000000007);
      (G[i][k] += G[i - 1][k]) %= (1000000007);
    }
  for (int k = 1; k <= M; k++) {
    long long S = 0;
    for (int i = N, j = 1; i >= 1; j++, i--) {
      (S += G[j][k]) %= (1000000007);
      (Ans += (F[i][k] * S) % (1000000007) * (M - k + 1)) %= (1000000007);
    }
  }
  if (Ans < 0) Ans += (1000000007);
}
void Output() { printf("%I64d\n", Ans); }
int main() {
  Init();
  Work();
  Output();
  return 0;
}
