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<stdio.h>
//#include <bits/stdc++.h>
#include<vector>
#include<float.h>
#include<iomanip>
#include<algorithm>
#include<string>
#include<cstring>
#include<math.h>
#include<cmath>
#include<sstream>
#include<set>
#include<map>
#include<queue>
#include <cassert>
#include <cmath>
#include<cstdint>

#define INF 1e9
#define rep(i,n)for(int i=0;(i)<(int)(n);i++)
#define REP(i,a,b)for(int i=(int)(a);(i)<=(int)(b);i++)
#define VEC(type, c, n) std::vector<type> c(n);for(auto& i:c)std::cin>>i;
#define vec(type,n) vector<type>(n)
#define vvec(m,n) vector<vector<int>> (int(m),vector<int>(n))
#define ALL(a)  (a).begin(),(a).end()
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)


using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;
using P = pair<ll, ll>;

int main(){
    int n, m, k;
    cin >> n >> m >> k;
    bool f = false;
    rep(i, m+1)if(n * i == k)f = true;
    rep(i, n+1)if(m * i == k)f = true;

    REP(i, 1, n-1){
        ll sum = m * i;
        REP(j, 1, m-1){
            sum = sum - i + n;
            if(sum == k)f = true;
        }
    }
    cout << (f ? "Yes" : "No") << endl;
}