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 <fstream>
#include <sstream>

#include <vector>
#include <set>
#include <bitset>
#include <map>
#include <deque>
#include <string>

#include <algorithm>
#include <numeric>

#include <cstdio>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>

#define pb push_back
#define pbk pop_back
#define mp make_pair
#define fs first
#define sc second
#define all(x) (x).begin(), (x).end()
#define foreach(i, a) for (__typeof((a).begin()) i = (a).begin(); i != (a).end(); ++i)
#define len(a) ((int) (a).size())

#ifdef CUTEBMAING
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define eprintf(...) 42
#endif

using namespace std;

typedef long long int64;
typedef long double ld;
typedef unsigned long long lint;

const int inf = (1 << 30) - 1;
const int64 linf = (1ll << 62) - 1;

int main() {
#ifdef XCODE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    int n, m, k; cin >> n >> m >> k;
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            int z = (n - i) * j + i * (m - j);
            if (z == k) {
                puts("Yes");
                return 0;
            }
        }
    }
    puts("No");
    return 0;
}