Codeforces 128 C. Games with Rectangle

长宽是各自独立的,算出各自可以取的方法数后相乘....

C. Games with Rectangle

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In this task Anna and Maria play the following game. Initially they have a checkered piece of paper with a painted n × m rectangle (only the border, no filling). Anna and Maria move in turns and Anna starts. During each move one should paint inside the last-painted rectangle a new lesser rectangle (along the grid lines). The new rectangle should have no common points with the previous one. Note that when we paint a rectangle, we always paint only the border, the rectangles aren't filled.

Nobody wins the game — Anna and Maria simply play until they have done k moves in total. Count the number of different ways to play this game.

Input

The first and only line contains three integers: n, m, k (1 ≤ n, m, k ≤ 1000).

Output

Print the single number — the number of the ways to play the game. As this number can be very big, print the value modulo 1000000007(109 + 7).

Sample test(s)

input

3 3 1

output

1

input

4 4 1

output

9

input

6 7 2

output

75

Note

Two ways to play the game are considered different if the final pictures are different. In other words, if one way contains a rectangle that is not contained in the other way.

In the first sample Anna, who performs her first and only move, has only one possible action plan — insert a 1 × 1 square inside the given3 × 3 square.

In the second sample Anna has as much as 9 variants: 4 ways to paint a 1 × 1 square, 2 ways to insert a 1 × 2 rectangle vertically, 2 more ways to insert it horizontally and one more way is to insert a 2 × 2 square.

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long int LL;

const LL MOD=1000000007LL;

LL n,m,k;
LL C[1200][1200];

void init()
{
    for(int i=0;i<1200;i++)
        C[i][i]=C[i][0]=1LL;
    for(int i=2;i<1200;i++)
    {
        for(int j=1;j<i;j++)
        {
            C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;
        }
    }
}

int main()
{
    init();
    cin>>n>>m>>k;
    if(n-1<2*k||m-1<2*k)
        cout<<0<<endl;
    else
        cout<<(C[n-1][2*k]*C[m-1][2*k])%MOD<<endl;
    return 0;
}