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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define FOR(n) for(int i=0;i<n;i++)
#define FORJ(n) for(int j=0;j<n;j++)
#define PRN(n) printf("%d\n",n)
#define PRF(n) printf("%lf\n",n)
#define PRL(n) printf("%lld\n",n)
#define PRS(s) printf("%s\n",s)
#define PRC(c) printf("%c",c)
#define INF 10000000000
#define mod 1000000007
typedef long long int ll;
ll u(const void *a, const void *b){return *(ll*)a-*(ll*)b;}
ll d(const void *a, const void *b){return *(ll*)b-*(ll*)a;}
ll min(ll a,ll b){
  if(a>b)return b;
  return a;
}
ll max(ll a,ll b){
  if(a>b)return a;
  return b;
}
ll gcd(ll a,ll b){
  if(!b)return a;
  return gcd(b,a%b);
}
ll lcm(ll a,ll b){return a*b/gcd(a,b);}
int kt(ll a){
  int sum=0;
  while(a){
    a/=10;
    sum++;
  }
  return sum;
}
int ks(ll a){
  int sum=0;
  while(a){
    sum+=a%10;
    a/=10;
  }
  return sum;
}
ll f(ll a){
  return a*(a+1)/2;
}
int main(void){
  int a,b,c;
  scanf("%d %d %d",&a,&b,&c);
  for(int i=1;i<=a;i++){
    for(int j=1;j<=b;j++){
      if(i*j+(a-i)*(b-j)==c){
        PRS("Yes");
        return 0;
      }
    }
  }
  if(c%a<1||c%b<1){
    PRS("Yes");
    return 0;
  }
  PRS("No");
  return 0;
}