Problem:
Score : 300 points

Problem Statement

Takahashi, Nakahashi and Hikuhashi have integers A, B and C, respectively.
After repeating the following operation K times, find the integer Takahashi will get minus the integer Nakahashi will get:

Each of them simultaneously calculate the sum of the integers that the other two people have, then replace his own integer with the result.

However, if the absolute value of the answer exceeds 10^{18}, print Unfair instead.

Constraints

1 \leq A,B,C \leq 10^9

0 \leq K \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C K

Output

Print the integer Takahashi will get minus the integer Nakahashi will get, after repeating the following operation K times.
If the absolute value of the answer exceeds 10^{18}, print Unfair instead.

Sample Input 1

1 2 3 1

Sample Output 1

1

After one operation, Takahashi, Nakahashi and Hikuhashi have 5, 4 and 3, respectively. We should print 5-4=1.

Sample Input 2

2 3 2 0

Sample Output 2

-1

Sample Input 3

1000000000 1000000000 1000000000 1000000000000000000

Sample Output 3

0

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

//set many funcs template
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<stdbool.h>
#include<time.h>
#define inf 1072114514
#define llinf 4154118101919364364
#define mod 1000000007
#define pi 3.1415926535897932384

int max(int a,int b){if(a>b){return a;}return b;}
int min(int a,int b){if(a<b){return a;}return b;}
int zt(int a,int b){return max(a,b)-min(a,b);}
int round(int a,int b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
int ceil(int a,int b){if(a%b==0){return a/b;}return (a/b)+1;}
int gcd(int a,int b){int c;while(b!=0){c=a%b;a=b;b=c;}return a;}
int lcm(int a,int b){int c=gcd(a,b);a/=c;return a*b;}
int nCr(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
int nHr(int a,int b){return nCr(a+b-1,b);}
int fact(int a){int i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
int pow(int a,int b){int i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
long long llmax(long long a,long long b){if(a>b){return a;}return b;}
long long llmin(long long a,long long b){if(a<b){return a;}return b;}
long long llzt(long long a,long long b){return llmax(a,b)-llmin(a,b);}
long long llround(long long a,long long b){if((a%b)*2 >= b){return (a/b)+1;}return a/b;}
long long llceil(long long a,long long b){if(a%b==0){return a/b;}return (a/b)+1;}
long long llgcd(long long a,long long b){long long c;while(b!=0){c=a%b;a=b;b=c;}return a;}
long long lllcm(long long a,long long b){long long c=llgcd(a,b);a/=c;return a*b;}
long long llnCr(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=(a+1-i);r/=i;}return r;}
long long llnHr(long long a,long long b){return llnCr(a+b-1,b);}
long long llfact(long long a){long long i,r=1;for(i=1;i<=a;i++){r*=i;}return r;}
long long llpow(long long a,long long b){long long i,r=1;for(i=1;i<=b;i++){r*=a;}return r;}
double dbmax(double a,double b){if(a>b){return a;}return b;}
double dbmin(double a,double b){if(a<b){return a;}return b;}
double dbzt(double a,double b){return dbmax(a,b)-dbmin(a,b);}
int sortfncsj(const void *a,const void *b){if(*(int *)a>*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int sortfnckj(const void *a,const void *b){if(*(int *)a<*(int *)b){return 1;}if(*(int *)a==*(int *)b){return 0;}return -1;}
int llsortfncsj(const void *a,const void *b){if(*(long long *)a>*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int llsortfnckj(const void *a,const void *b){if(*(long long *)a<*(long long *)b){return 1;}if(*(long long *)a==*(long long *)b){return 0;}return -1;}
int dbsortfncsj(const void *a,const void *b){if(*(double *)a>*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int dbsortfnckj(const void *a,const void *b){if(*(double *)a<*(double *)b){return 1;}if(*(double *)a==*(double *)b){return 0;}return -1;}
int strsortfncsj(const void *a,const void *b){return strcmp((char *)a,(char *)b);}
int strsortfnckj(const void *a,const void *b){return strcmp((char *)b,(char *)a);}

int main(void){
    long long i,j,n,m,k,a,b,c,w,r=0,l,t;
    double d;
    char s[262144];
    scanf("%lld%lld%lld%lld",&a,&b,&c,&k);
    //l=strlen(s);
    //for(i=0;i<n;i++){scanf("%d",&a[i]);}
    //qsort(a,n,sizeof(int),sortfncsj);
    if(k%2){printf("%lld\n",b-a);}else{printf("%lld\n",a-b);}
    return 0;
}