Problem:
Score: 200 points

Problem Statement

There are three positive integers A, B and C written on a blackboard. E869120 performs the following operation K times:

Choose one integer written on the blackboard and let the chosen integer be n. Replace the chosen integer with 2n.

What is the largest possible sum of the integers written on the blackboard after K operations?

Constraints

A, B and C are integers between 1 and 50 (inclusive).

K is an integer between 1 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:

A B C
K

Output

Print the largest possible sum of the integers written on the blackboard after K operations by E869220.

Sample Input 1

5 3 11
1

Sample Output 1

30

In this sample, 5, 3, 11 are initially written on the blackboard, and E869120 can perform the operation once.

There are three choices:

Double 5: The integers written on the board after the operation are 10, 3, 11.

Double 3: The integers written on the board after the operation are 5, 6, 11.

Double 11: The integers written on the board after the operation are 5, 3, 22.

If he chooses 3., the sum of the integers written on the board afterwards is 5 + 3 + 22 = 30, which is the largest among 1. through 3.

Sample Input 2

3 3 4
2

Sample Output 2

22

E869120 can perform the operation twice. The sum of the integers eventually written on the blackboard is maximized as follows:

First, double 4. The integers written on the board are now 3, 3, 8.

Next, double 8. The integers written on the board are now 3, 3, 16.

Then, the sum of the integers eventually written on the blackboard is 3 + 3 + 16 = 22.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, start, end) for (long long i = start; i < end; ++i)
#define repreverse(i, start, end) for (long long i = start; i >= end; --i)
#define all(x) (x).begin(), (x).end()
#define len(x) ((long long)(x).size())
#define lcm(a, b) ((a) / __gcd((a), (b)) * (b))
using namespace std;
using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vllvll = vector<vll>;
using pll = pair<ll, ll>;
template<class T>void print1d(T x,ll n=-1){if(n==-1)n=x.size();rep(i,0,n){cout<<x[i]<<' ';}cout<<'\n';}
template<class T>void print2d(T x,ll r=-1,ll c=-1){if(r==-1)r=x.size();if(c==-1)c=x[0].size();rep(i,0,r)print1d(x[i],c);}
template<class T, class U>bool haskey(T mp, U key) { return mp.find(key) != mp.end(); }
template<class T, class U>bool isin(T el, U container) { return find(all(container), el) != container.end(); }
template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }
template<class T>bool even(T n) { return ! (n & 1); }
template<class T>bool odd(T n) { return n & 1; }
template<class T>ld deg2rad(T deg) { return M_PI * deg / 180.0; }
template<class T>ld rad2deg(T rad) { return 180.0 * rad / M_PI; }
ll intpow(ll a,ll n){ll p=1;while(n){if(n&1)p*=a;a*=a;n>>=1;}return p;}
const long double pi = M_PI;
const long long big = 1LL << 50;
const long long inf = 1LL << 60;
const long long mod = 1e9 + 7;



int main()
{
        vll nums(3);
        cin >> nums[0] >> nums[1] >> nums[2];
        ll K;
        cin >> K;
        rep(i, 0, K) {
                *max_element(all(nums)) *= 2;
        }
        cout << accumulate(all(nums), 0LL) << endl;
}