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
/* 2013-07-26 09:26:54.455070 */#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <iostream>
using namespace std;

const long long mod = 1000000007;
const int MAXN = 2005;
int n, m;
long long f[MAXN][MAXN], g[MAXN][MAXN];

int main() {

	scanf("%d %d", &n, &m);

	for(int i = 2; i <= m; i++)
		f[1][i] = 1;
	for(int i = 2; i <= n; i++) {
		long long sum = 0, summ = 0;
		for(int j = 2; j <= m; j++) {
			sum = (sum + f[i - 1][j]) % mod;
			summ = (summ + f[i - 1][j] * j % mod) % mod;
			f[i][j] = (-summ + sum * (j + 1) % mod + mod) % mod; 
		}
	}
	for(int i = 2; i <= m; i++)
		g[0][i] = 1;
	for(int i = 1; i <= n; i++) {
		long long sum = 0, summ = 0;
		for(int j = 1; j <= m; j++) {
			g[i][j] = (-summ + sum * (j + 1) % mod + mod) % mod;
			sum = (sum + f[i][j]) % mod;
			summ = (summ + f[i][j] * j % mod) % mod;
		}
	}
	long long ans = 0;
	for(int i = 2; i <= m; i++) {
		long long sum1 = 0;
		for(int j = 1; j <= n; j++) {
			sum1 = (sum1 + f[j][i]) % mod;
		}
		long long sum2 = g[0][i], res = 0;
		for(int j = n; j >= 1; j--) {
			res = (res + sum1 * sum2 % mod) % mod;
			sum2 = (sum2 + g[n - j + 1][i]) % mod;
			sum1 = (sum1 - f[j][i] + mod) % mod;
		}
		ans = (ans + res * (m - i + 1) % mod) % mod;
	}
	cout << ans << endl;
	return 0;
}
