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 (GCC 5.4.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS

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

//for
#define REP(i,n) for(int i=0,i##_len=(int)(n);i<i##_len;i++)
#define REPS(i,n) for(int i=1,i##_len=(int)(n);i<=i##_len;i++)
#define RREP(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define RREPS(i,n) for(int i=(int)(n);i>0;i--)
#define SREP(i,s,n) for(int i=(int)(s),i##_len=(int)(n);i<i##_len;i++)

//printf
#define PRINTD(d) printf("%d\n",(d))

//memset
#define m0(s) memset(s,0,sizeof(s))
#define ml(s) memset(s,63,sizeof(s))
#define fill(s,c) memset(s,c,sizeof(s))

#define INF 1e9

typedef long long ll;
int diff[4][2] = {
	{ 0, -1 },
	{ -1, 0 },
	{ 1, 0 },
	{ 0, 1 },
};

int Min(int a, int b) {
	return (a)<(b) ? (a) : (b);
}
int Max(int a, int b) {
	return (a) > (b) ? (a) : (b);
}
void Swap(int *a, int *b) {
	int tmp = *a;
	*a = *b;
	*b = tmp;
}

int main() {
	int a, b, c;
	scanf("%d %d %d", &a, &b, &c);
	if (b < c)Swap(&b, &c);
	if (a < b)Swap(&a, &b);
	int cnt = 0;
	if ((a - b) % 2 && (a - c) % 2) {
		b++;
		c++;
		cnt++;
	}
	else if ((a - b) % 2) {
		a++;
		c++;
		cnt++;
	}
	else if ((a - c) % 2) {
		a++;
		b++;
		cnt++;
	}
	cnt += (a - b) / 2 + (a - c) / 2;
	PRINTD(cnt);
}