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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> II;
typedef pair<long long, int> LLI;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef vector<II> VII;

#define For(i,a,b) for(int i = a;i <= b; i++)
#define Rep(i,a,b) for(int i = a;i >= b; i--)
#define REP(i, n) for(int i = 0; i < n; i++)
#define FOR(i, f) for(auto i : f)
#define fi first
#define se second
#define pb push_back
#define sz(s) int(s.size())
#define reset(f, x) memset(f, x, sizeof(f))
#define all(x) x.begin(), x.end()
#define two(x) (1LL << x)
#define getbit(x, i) ((x >> (i-1)) & 1LL)
#define onbit(x, i) (x | (1LL << (i-1)))
#define offbit(x, i) (x & ~(1 << (i-1)))

int main() {
    ios_base::sync_with_stdio(false);
    //freopen("in.txt","r",stdin);
    //freopen("","w",stdout);
    LL a, b, c, k;
    cin >> a >> b >> c >> k;
    if (k % 2) cout << b-a; else cout << a-b;
}