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<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#include<iomanip>
typedef long long int ll;
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define REP(i,n) for (int i=0;i<(n);i++)
#define EREP(i,n) for (int i=1;i<=(n);i++)
#define ALL(a) (a).begin(),(a).end()
//#define EVEL 1

#ifdef EVEL
#define DEB(X) cout << #X << ":" <<X<<" " ;
#define TF(f) f ? cout<<"true  " : cout<<"false ";
#define END cout<<"\n";
#else
#define DEB(X) {}
#define TF(f) {}
#define END {}
#endif
const int MOD = 1000000007;
const int INF = 2000000000;

    int N,M,K;
    bool F=false;

int main(){
    ios_base::sync_with_stdio(false);
    cin>>N>>M>>K;
    if(N>M)swap(N,M);
    EREP(i,N){
        EREP(j,M){
            if(M*i-(j*i)+j*(N-i)==K){
                F=true;
            }
            if(F)break;
        }
        if(F)break;

    }
    cout<<((F||K==0||K==N*M)?"Yes":"No");
    return 0;
}