Problem:
Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23

Solution in C (Clang 3.8.0) (judge verdict: Runtime Error)

#include <stdio.h>
#define max(a, b) ((a) > (b) ? (a) : (b))

int
main(int argc, char *argv[])
{
  int a[3];
  scanf("%d %d %d", &a[0], &a[1], &a[2]);

  int m = max(a[0], max(a[1], a[2]));

  int cnt = 0; int k[2]; int j = 0;
  for(int i = 0; i < 3; i++){
    if(m - a[i] != 0){
      k[j++] = i; 
      while(m > a[i]){
	cnt++;
	a[i] += 2;
      }
      if(a[i] - m == 1){
	a[i] -= 2;
	cnt--;
      }
    }
  }

  if(((m - a[k[0]]) ^ (m - a[k[1]]))) cnt += 2;
  if(((m - a[k[0]]) & (m - a[k[1]]))) cnt++;
  
  
  printf("%d\n", cnt);

  return 0;
}