CodeForces 368A

Time Limit:1000MS    Memory Limit:262144KB    64bit IO Format:%I64d
& %I64u

Description

Sereja owns a restaurant for n people. The restaurant hall has a coat rack withn hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using thei-th
hook costs ai rubles. Only one person can hang clothes on one hook.

Tonight Sereja expects m guests in the restaurant. Naturally, each guest wants to hang his clothes on an available hook with minimum price (if there are multiple such hooks, he chooses any of them). However if the moment
a guest arrives the rack has no available hooks, Sereja must pay a
d ruble fine to the guest.

Help Sereja find out the profit in rubles (possibly negative) that he will get tonight. You can assume that before the guests arrive, all hooks on the rack are available, all guests come at different time, nobody besides them
guests is visiting Sereja's restaurant tonight.

Input

The first line contains two integers n andd(1 ≤ n, d ≤ 100). The next line contains integersa1,
a2,
...,an(1 ≤ ai ≤ 100). The third line contains integerm(1 ≤ m ≤ 100).

Output

In a single line print a single integer — the answer to the problem.

Sample Input

Input

2 1
2 1
2

Output

3

Input

2 1
2 1
10

Output

-5

Hint

In the first test both hooks will be used, so Sereja gets
1 + 2 = 3 rubles.

In the second test both hooks will be used but Sereja pays a fine
8 times, so the answer is 3 - 8 =  - 5.

#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int m,n,d,a[111],i,ans=0;
cin>>n>>d;
for(i=0;i<n;i++)
{
cin>>a[i];
}
sort(a,a+n);
cin>>m;
if(m<=n)
{
for(i=0;i<m;i++)
{
ans+=a[i];
}
}
else
{
for(i=0;i<n;i++)
{
ans+=a[i];
}
ans=ans-(m-n)*d;
}
cout<<ans<<endl;
return 0;
}