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;
template <typename T, typename L>
bool smax(T &x, L y) {
  return x < y ? (x = y, true) : false;
}
template <typename T, typename L>
bool smin(T &x, L y) {
  return y < x ? (x = y, true) : false;
}
const long long MAXN = 2048;
const long long mod = 1e9 + 7;
long long n, m;
long long c[MAXN << 1][MAXN << 1];
long long dp[MAXN][MAXN];
long long ans;
int32_t main() {
  cin >> n >> m;
  for (long long i = 0; i < MAXN; i++)
    for (long long j = 0; j <= i; j++) {
      if (j == 0) {
        c[i][j] = 1;
        continue;
      }
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
    }
  for (long long w = 1; w <= m; w++)
    for (long long h = 0; h < n; h++)
      dp[w][h] = (c[(h << 1) + (w - 1)][w - 1] + (h ? dp[w][h - 1] : 0)) % mod;
  for (long long w = 1; w <= m; w++)
    for (long long h = 0; h < n; h++)
      ans = (ans + ((long long)((m - w) * c[(h << 1) + (w - 1)][w - 1]) % mod) *
                       (long long)(dp[w][n - h - 1])) %
            mod;
  return cout << ans << '\n', false;
}
