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 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>


void swap (int *x, int *y) {
  int temp;

  temp = *x;
  *x = *y;
  *y = temp;

  return;
}

int gcd(int a,int b){
  if(a<b){
    swap(&a,&b);
  }

  if(b==0){
    return a;
  }

  int r=a%b;
  while(r!=0){
    a = b;
    b = r;
    r = (a%b);
  }

  return b;
}

/* クイックソート */
int dn(const void*a,const void*b){return*(int*)b-*(int*)a;}


long int modpow(long int a,long int n,long int mod){
  long int ans=1;
  while(n>0){
    if(n & 1){
      ans=ans*a%mod;
    }
    a=a*a%mod;
    n/=2;
  }
  return ans;
}

int max(int a,int b){
  if(a<b){
    return b;
  }
  else{
    return a;
  }
}

int min(int a,int b){
  if(a<b){
    return a;
  }
  else{
    return b;
  }
}

void chmax(int *a,int b){
  if(*a < b){
    *a = b;
  }

  return;
}

void chmin(int *a,int b){
  if(*a > b){
    *a =b;
  }

  return;
}

/****************************************\
| Thank you for viewing my code:)        |
| Written by RedSpica a.k.a. RanseMirage |
| Twitter:@asakaakasaka                  | 
\****************************************/

int main(void){
  int h,w,k;
  scanf("%d%d%d",&h,&w,&k);

  int ans=0;
  for(int i=0;i<=h;i++){
    for(int j=0;j<=w;j++){
      int now=(w*i)+(h*j)-2*i*j;

      if(now==k){
        ans=1;
      }
    }
  }

  if(ans==1){
    printf("Yes\n");
  }
  else{
    printf("No\n");
  }

  return 0;
}