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 09:47:35.567940 */#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cassert>
#include <sstream>
#include <bitset>
#include <numeric>
#include <climits>
#include <string>
#include <cctype>
#include <ctime>
#include <iomanip>
#include <cmath>
#include <vector>
#include <queue>
#include <list>
#include <map>
#include <set>
using namespace std;

#define foreach(e, x) for (__typeof((x).begin()) e = (x).begin(); e != (x).end(); ++e)

typedef long long int64; 

const int N = 2222;
const int MOD = int(1e9) + 7;

int n, m;
int f[N][N];
int g[N][N];
int h[N][N];
int comb[N][N];

inline void add(int &x, int t) {
	if ((x += t) >= MOD) {
		x -= MOD;
	}
}

int main() {
	for (int i = 0; i < N; i++) {
		comb[i][0] = 1;
		for (int j = 1; j <= i; j++) {
			comb[i][j] = comb[i - 1][j - 1] + comb[i - 1][j];
			if (comb[i][j] >= MOD) {
				comb[i][j] -= MOD;
			}
		}
	}

	scanf("%d%d", &n, &m);
	for (int j = 2; j <= m; j++) {
		g[1][j] = f[1][j] = 1;
		add(g[1][j], g[1][j - 1]);
	}
	for (int i = 2; i <= n; i++) {
		int now = 0;
		for (int j = 1; j <= m; j++) {
			add(now, g[i - 1][j]);
			g[i][j] = f[i][j] = now;
			add(g[i][j], g[i][j - 1]);
		}
	}
	for (int j = 1; j <= m; j++) {
		for (int i = 1; i <= n; i++) {
			h[i][j] = f[i][j];
			add(h[i][j], h[i - 1][j]);
		}
	}
	memcpy(g, h, sizeof(g));
	memset(h, 0, sizeof(h));
	for (int i = 1; i <= n; i++) {
		int now = 0, tmp = 0;
		for (int j = 1; j <= m; j++) {
			add(now, tmp);
			h[i][j] = now;
			add(now, g[i][j]);
			add(tmp, g[i][j]);
		}
	}
	int result = 0;
	for (int row = 1; row <= n; row++) {
		for (int length = 1; length <= m; length++) {
			int now = (h[row - 1][length] + 1) % MOD;
			now = 1LL * now * g[n - row + 1][length] % MOD;
			now = 1LL * now * (m - length + 1) % MOD;
			add(result, now);
		}
	}
	cout << result << endl;
}
