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 <bits/stdc++.h>
#include <iomanip>
#include <math.h>
#define rep(i,x) for(ll i = 0;i<x;i++)
#define all(a) (a).begin(),(a).end()
using ll = long long;
using ld = long double;
using namespace std;
const ll INF = 1001001001;
const ll mod = 1000000007;
typedef pair<ll,ll> P;
using graph = vector<vector<ll>>;
ll gcd(ll a,ll b){
    if(a < b)swap(a,b);
    if(a % b == 0)return b;
    else return gcd(b,a%b);
}
bool isp(ll n){
    bool res = true;
    if(n == 1)return false;
    else{
        for(ll i = 2;i * i <= n;i++){
            if(n % i == 0){
                res = false;
                break;
            }
        }
        return res;
    }
}
//printf("%.10f\n", N);
/*
ll memo[100000];
ll fibo(ll n){
    if(memo[n] != 0)return memo[n];
    if(n <= 1)return n;
    else return memo[n] = fibo(n - 1)+ fibo(n - 2);
}*/

int main(){
    ll N,M;
    ll K;
    cin >> N >> M >> K;
    bool ans = false;
    rep(i,N + 1){
        rep(j,M + 1){
            if(j * (N - i) + i * (M - j) == K)ans = true;
        }
    }
    if(ans)cout << "Yes" << endl;
    else cout << "No" << endl;
}