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
#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <queue>
#define int long long int

using namespace std;


main()
{
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    int N, M;
    int mod = 1e9+7;
    cin>>N>>M;
    vector<vector<int> > D(N, vector<int>(M+1));
    vector<vector<int> > Sum(N, vector<int>(M+1));
    vector<vector<int> > Sum2(N, vector<int>(M+1));
    for(int len = 2; len <= M; ++ len)
    {
        D[0][len] = 1;
        Sum[0][len] = (Sum[0][len-1] + D[0][len]) % mod;
        Sum2[0][len] = (Sum2[0][len-1] + Sum[0][len]) % mod;
    }
    for(int i = 1; i < N; ++ i)
    {
        for(int len = 2; len <= M; ++ len)
        {
            D[i][len] = (1 + Sum2[i-1][len]) % mod;
            Sum[i][len] = (Sum[i][len-1] + D[i][len]) % mod;
            Sum2[i][len] = (Sum2[i][len-1] + Sum[i][len]) % mod;
            //cout<<i<<' '<<len<<' '<<D[i][len]<<endl;
        }
    }
    int ans = 0;
    int i = 0;
    for(int len = 2; len <= M; ++ len)
    {
        ans = (ans + ((M-len+1) * D[N-1-i][len]) % mod) % mod;
    }
    //cout<<ans<<endl;
    for(int i = 1; i < N; ++ i)
    {
        for(int len = 2; len <= M; ++ len)
        {
            ans = (ans + ((M-len+1) * ((D[i][len] - D[i-1][len]) * D[N-1-i][len]) % mod) % mod) % mod;
        }
    }
    cout<<ans<<endl;
}
