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 11:05:35.465811 */#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>

#define MM(a, b) memset(a, b, sizeof(a))
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;

const int maxn = 2111;
const int maxm = 111;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;

int n, m;

void solve(){}

const long long mod = 1000000007;

long long F[maxn][maxn];
long long G[maxn][maxn];

int main(){
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i){
        long long tot = 0;
        for (int j = 2; j <= m; j++){
            F[i][j] = (1 + G[i - 1][j]) % mod;
            tot = (tot + F[i][j]) % mod;
            G[i][j] = (G[i][j - 1] + tot) % mod;
        }
    }
    long long ret = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 2; j <= m; j++){
            ret = (ret + F[i][j] * ((G[n - i][j] - F[n - i][j] + 1ll) % mod + mod) % mod * (m - j + 1) % mod) % mod;
        }
    cout<<ret<<endl;
}
