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;
clock_t __START, __END;
double __TOTALTIME;
void _MS() { __START = clock(); }
void _ME() {
  __END = clock();
  __TOTALTIME = (double)(__END - __START) / CLOCKS_PER_SEC;
  cout << "Time: " << __TOTALTIME << " s" << endl;
}
const double E = 2.718281828;
const double PI = acos(-1.0);
const long long INF = (1LL << 60);
const int inf = (1 << 30);
const double ESP = 1e-9;
const int mod = (int)1e9 + 7;
const int N = (int)2e3 + 10;
long long sq[N][N];
long long sum[N];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  if (m < 2) {
    printf("0\n");
    return 0;
  }
  for (int i = 2; i <= m; ++i) sq[i][1] = 1;
  for (int i = 2; i <= n; ++i) {
    sq[2][i] = 1;
    sum[2] = 1;
    for (int j = 3; j <= m; ++j)
      sum[j] = sq[j][i - 1] + sum[j - 1], sum[j] %= mod;
    long long temp = 1;
    for (int j = 3; j <= m; ++j) temp += sum[j], temp %= mod, sq[j][i] = temp;
  }
  for (int j = 2; j <= m; ++j)
    for (int i = 2; i <= n; ++i) {
      sq[j][i] += sq[j][i - 1];
      sq[j][i] %= mod;
    }
  long long ans = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 2; j <= m; ++j) sum[j] = sq[j][n - i];
    for (int j = 3; j <= m; ++j) sum[j] += sum[j - 1], sum[j] %= mod;
    ans += sq[2][i] * (m - 2 + 1) % mod;
    ans %= mod;
    long long temp = 0;
    for (int j = 3; j <= m; ++j)
      ans += sq[j][i] * (m - j + 1) % mod, ans %= mod;
    for (int j = 3; j <= m; ++j) {
      temp += sum[j - 1];
      temp %= mod;
      temp += sq[j - 1][n - i];
      temp %= mod;
      ans += sq[j][i] * temp % mod * (m - j + 1) % mod;
      ans %= mod;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
