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 10:54:21.988245 */#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <ctime>
#include <sstream>
#include <fstream>

using namespace std;

typedef long long int64;

#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define ROF(i,a,b) for(int i=(a);i>=(b);--i)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define sz(a) ((int)(a).size())
#define ms(a,x)

const int mo = 1000000007;
const int N = 2005;

long long f[N][N], g[N][N], s[N][N];
bool v[N][N];

long long dp(int i, int j)
{
	if (j < 2) return 0;
	if (v[i][j]) return s[i][j];
	if (i == 1)
	{
		f[i][j] = 1;
		g[i][j] = j - 1;
		s[i][j] = 1;
		return s[i][j];
	}
	v[i][j] = true;
	dp(i - 1, j);
	dp(i, j - 1);
	f[i][j] = (f[i][j - 1] + g[i - 1][j - 1] + f[i - 1][j]) % mo;
	g[i][j] = (g[i][j - 1] + f[i][j]) % mo;
	s[i][j] = (f[i][j] + s[i-1][j]) %mo;
	return s[i][j];
}

int main()
{
	#ifdef LOCAL_TEST
		//freopen("a.in","r",stdin);
		//freopen("a.out","w",stdout);
	#endif
	int n, m;
	cin >> n >> m;
	long long ans = 0;
	for (int i = 1; i <= n; ++i)
		for (int j = 2; j <= m; ++j)
		{
			long long F1 = dp(i, j);
			if (i != 1)
			{
				F1 -= dp(i-1, j);
				F1 += mo;
				F1 %= mo;
			}
			long long F2 = dp(n - i + 1, j);
			ans += F1 * F2 % mo * (m - j + 1) % mo;
			ans %= mo;
		}
	cout << ans << endl;
	return 0;
}
