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: Time Limit Exceeded)

#include <iostream>
#include <vector>
#include <queue>
#include <sstream>
#include <algorithm>
#include <bitset>
#include <limits>
#include <map>
#include <set>
#include <iomanip>
#include <cmath>

using namespace std;

typedef long long int ll;
typedef std::numeric_limits<double> dbl;
const long long int LL_INF=1LL<<60;

vector<string> split(const string& input, char delimiter)
{
    stringstream stream(input);
    
    string field;
    vector<string> result;
    while (getline(stream, field, delimiter)) {
        result.push_back(field);
    }
    return result;
}

//#define DEBUG

ll A, B, M, N, D, K, L, Q, W, H, T, X, Y, S;

bool field[1000][1000];

int main()
{
    cin >> N >> M >> K;
    
    for (int n = 0; n <= N; ++n) {
        for (int m = 0; m <= M; ++m) {
            ll cnt = 0;
            cnt += n * M;
            cnt += m * N;
            cnt -= 2*n*m;
            //cnt -= (M-m) * (N-n);
            cerr << n << " " << m << " " << cnt << endl;
            if (cnt == K) {
                cout << "Yes" << endl;
                return 0;
            }
        }
    }

    cout << "No" << endl;

    return 0;
}