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: Wrong Answer)

#include <iostream>
#include <float.h>
#include <vector>
#include <algorithm>
#include <string>
#include <map>
#include <iomanip>
#include <sstream>
#include <bitset>
#include <fstream>
#include <queue>
#include <set>
#include <stdlib.h>
#include <time.h>
#include <list>
#include <stack>
#include <queue>
#include <sstream>

#define For(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  For(i,0,n)

//clear memory
#define CLR(a) memset((a), 0 ,sizeof(a))
#define SORT(c) sort((c).begin(),(c).end())
#define vi(m,a) vector<int> m(a)
#define vti(m,a,i) vector<vector<int>> m(a,vector<int>(i))
#define ALL(it,a) for(auto it = a.begin(); it!=a.end(); it++)
#define Fe(it,a) for(auto &it : a)
#define all(a) begin(a),end(a)

using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef pair<ll, ll> pll;
#define WARU 1000000007;

int dx[] = { 1,0,-1,0 };
int dy[] = { 0,1,0,-1 };

int main() {
	int n, m;
	int k;
	cin >> n >> m >> k;
	bool flag = false;
	rep(i,n)
		rep(l, m) {
		if (n*i + m *l - 2 * i*l == k)
			flag = true;

	}
	if (flag)
		cout << "Yes" << endl;
	else
		cout << "No" << endl;


}