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: Wrong Answer)

#include <bits/stdc++.h>
#include<cmath>
#define N 100005
#define A 1000005
#define MOD 998244353 //1000000007 // //100000000
#define INF 1000000000000000000
#define ll long long
//#define M_PI 3.141592653589793238463

using namespace std;

#define pii pair<ll, ll>
#define piii pair<ll, pii>
#define ft first
#define sd second
#define pb push_back
#define rep(i, n) for(ll i = 0; i < n; i++)
#define repr(i, n) for(ll i = n-1; i >= 0; i--)
#define itr(it, x) for(auto it = x.begin(); it != x.end(); it++)
#define mem(a, b) memset(a, (ll)b, sizeof(a))
#define all(v) v.begin(), v.end()
#define allr(v) v.rbegin(), v.rend()
#define edge(v, x, y) v[x].pb(y); v[y].pb(x);
#define popcount __builtin_popcount
#define ANS(s) {cout << s << "\n"; return;}

//~~~~~~fancy print~~~~~~~//
#define printpii(a) cout << a.ft << " " << a.sd << endl;
#define printpiii(a) cout << a.ft << " " << a.sd.ft << " " << a.sd.sd << endl;
#define print(a, n) rep(i, n) cout << a[i] << " "; cout << endl;
#define printv(v) for(auto x: v)cout << x << " "; cout << endl;
#define printm(a, n, m) rep(i, n) { rep(j, m) cout << a[i][j] << "\t"; cout << endl;}

ll dx[4] = {0, 0, 1, -1};
ll dy[4] = {1, -1, 0, 0};

void fast(){ios_base::sync_with_stdio(false);cin.tie(0);cout << setprecision(12) << fixed;}


void solve(){
    ll a, b, c, k;
    cin >> a >> b >> k;
    cout << (k%2 == 1? b-a: a-b);
}
main(){
    /*#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif*/
    fast();
    ll t = 1; //cin >> t;
    while(t--){
        solve();
    }
}