Problem:
Score : 200 points

Problem Statement

We have a grid with N rows and M columns of squares. Initially, all the squares are white.

There is a button attached to each row and each column.
When a button attached to a row is pressed, the colors of all the squares in that row are inverted; that is, white squares become black and vice versa.
When a button attached to a column is pressed, the colors of all the squares in that column are inverted.

Takahashi can freely press the buttons any number of times. Determine whether he can have exactly K black squares in the grid.

Constraints

1 \leq N,M \leq 1000

0 \leq K \leq NM

Input

Input is given from Standard Input in the following format:

N M K

Output

If Takahashi can have exactly K black squares in the grid, print Yes; otherwise, print No.

Sample Input 1

2 2 2

Sample Output 1

Yes

Press the buttons in the order of the first row, the first column.

Sample Input 2

2 2 1

Sample Output 2

No

Sample Input 3

3 5 8

Sample Output 3

Yes

Press the buttons in the order of the first column, third column, second row, fifth column.

Sample Input 4

7 9 20

Sample Output 4

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

//C++14 (GCC 5.4.1)

#define LLINF (1LL<<60)
typedef long long ll;
#define mod 1000000007

#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
#define rrepd(i,a,b) for(int i=(a);i>=(b);i--)

template<typename T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<typename T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

template<typename T> inline T divrup(T a, T b) { if (a % b) return a/b+1; else return a/b;}

#define dbg_v1(i,V) do{cout<<"-v1----\n";rep(i,V.size())cout<<" "<<i<<"\t:"<<V[i]<<endl;cout<<endl;}while(0)
#define dbg_v2(i,k,V) do{cout<<"-v2----\n";rep(i,V.size()){rep(k,V[0].size())cout<<" "<<V[i][k];cout<<endl;}}while(0)

int main() {
    ios::sync_with_stdio(false);//stdoutとcoutの同期解除
    cin.tie(nullptr);//cinとcoutの同期解除
    
    int N,M,K;
    cin >>N>>M>>K;

    //-----------------------------------
    bool is=false;

    rep(i,N+1)rep(k,M+1){
        if(i*(M-k)+(N-i)*k == K){
            is=true;
            goto F;
        }
    }
F:
    cout << ((is) ? "Yes" : "No") <<endl;
    return (0);
}