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)

/*
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*/
#include "bits/stdc++.h"
using namespace std;
#define ok1 printf("ok1\n");
#define ok2 printf("ok2\n");
#define M 1000000000000000000LL
#define rep(i,n) for(int i=0;i<n;++i)
#define REP(i,s,n) for(int i=(s);i<(n);++i)
#define repr(i,n) for(int i=n-1;i>=0;--i)
#define REPR(i,s,n) for(int i=(s);i>=(n);--(i))
#define all(a) (a).begin(),(a).end()
#define reall(a) (a).rbegin(),(a).rend()
#define pb push_back
#define pf push_front
#define MIN(a,b) a=min((a),(b))
#define MAX(a,b) a=max((a),(b))
#define SIZE(v) (int)v.size()
#define DOUBLE fixed << setprecision(15)
#define fi first
#define se second
#define mp make_pair
const double pi = acos(-1.0);
typedef vector<int> vi;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<char> vc;
typedef vector<double> vd;
typedef vector<bool> vb;
typedef deque<ll> dll;
typedef pair<ll, ll> P;
typedef vector<P> vP;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
ll dy[4] = { 1,0,-1,0 };
ll dx[4] = { 0,1,0,-1 };
struct edge {
	int to;
	int cost;
};
bool out_check(ll a, ll b) {
	if (0 <= a && a < b) return true;
	else return false;
}
void Pvll(vll v) {
	cout << "vll------------------------------------------------\n";
	rep(i, v.size()) cout << v[i] << " ";
	cout << endl;
	cout << "------------------------------------------------\n";
}
void Pvvll(vvll v) {
	cout << "vvll------------------------------------------------\n";
	rep(i, v.size()) {
		rep(j, v[i].size()) {
			cout << v[i][j] << " ";
		}
		cout << endl;
	}
	cout << "------------------------------------------------\n";
}

void Ps(string s) {
	cout << "str------------------------------------------------\n";
	rep(i, s.size()) cout << s[i] << " ";
	cout << endl;
	cout << "------------------------------------------------\n";
}

void Pvs(vs s) {
	cout << "vs------------------------------------------------\n";
	rep(i, s.size()) {
		rep(j, s[i].size()) {
			cout << s[i][j] << " ";
		}
		cout << endl;
	}
	cout << "------------------------------------------------\n";
}

void Yes(bool x) {
	if (x) cout << "Yes\n";
	else cout << "No\n";
}

void YES(bool x) {
	if (x) cout << "YES\n";
	else cout << "NO\n";
}

void yes(bool x) {
	if (x) cout << "yes\n";
	else cout << "no\n";
}

void Yay(bool x) {
	if (x) cout << "Yay!\n";
	else cout << ":(\n";
}

ll  n, m, d, r, l, k, h, ans, ret = M;
bool flag = false, flag2 = false, flag3 = false;
string s, t, u;

void INIT() {
	cin.tie(0);
	ios::sync_with_stdio(false);
}

void addmod(ll &a, ll &b) {
	a = (a + b) % mod;
}

vll divisor(ll n)//nの約数を列挙
{
	vll w;
	for (int i = 1; i*i <= n; ++i)
	{
		if (n%i == 0)
		{
			w.push_back(i);
			//cout << i << endl;
			if (i*i != n)
			{
				w.push_back(n / i);
				//cout << n/i << endl;
			}
		}
	}
	return w;
}

int main() {
	cin >> n >> m >> k;
	rep(i, n + 1)rep(j, m + 1) {
		if (i * j + (n - i)*(m - j) == k) {
			Yes(1);
			return 0;
		}
	}
	Yes(0);
	return 0;
}