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 11:25:58.948397 */#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <cmath>

typedef long long LL;
#define pb push_back
using namespace std;

template<class T> T abs(T x){if (x < 0) return -x; else return x;}

const int maxn = 2100, mo = 1000000007;
LL f[maxn][maxn], f2[maxn][maxn], s[maxn], h[maxn];
int n, m;

int main(){
	scanf("%d%d", &n, &m);
	memset(f, 0, sizeof(f));
	memset(f2, 0, sizeof(f2));
	memset(h, 0, sizeof(h));
	memset(s, 0, sizeof(s));
	for (int i = 2; i <= m; ++i){
		f[1][i] = 1;
		f2[1][i] = 1;
		s[i] = i - 1;
		h[i] = (h[i - 1] * (LL)i) % mo;
	}
	for (int i = 2; i <= n; ++i){
		for (int j = 2; j <= m; ++j){
			f[i][j] = (f[i - 1][j] + (s[j - 1] * (LL)(j + 1) - h[j - 1])) % mo;
			f2[i][j] = ((s[j - 1] * (LL)(j + 1) - h[j - 1])) % mo;
		}
		memset(s, 0, sizeof(s));
		memset(h, 0, sizeof(h));
		for (int j = 2; j <= m; ++j){
			s[j] = (s[j - 1] + f[i][j]) % mo;
			h[j] = (h[j - 1] + f[i][j] * (LL)j) % mo;
		}
	}

	LL ans = 0;
	for (int L = 1; L <= n; ++L){
		LL sum = 0;
		for (int j = 2; j <= m; ++j)
			sum = (sum + f2[L][j] * f[n - L + 1][j] * (LL)(m - j + 1)) % mo;
		ans = (ans + sum) % mo;
	}
	printf("%I64d\n", ans);
	return 0;
}
