Sereja owns a restaurant for n people. The restaurant hall has a coat rack with n hooks. Each restaurant visitor can use a hook to hang his clothes on it. Using the i-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 the m guests is visiting Sereja's restaurant tonight.
The first line contains two integers n and d (1 ≤ n, d ≤ 100). The next line contains integers a1, a2, ..., an (1 ≤ ai ≤ 100). The third line contains integer m (1 ≤ m ≤ 100).
In a single line print a single integer — the answer to the problem.
2 1 2 1 2
3
2 1 2 1 10
-5
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.
题意说是有个叫Sereja的屌丝开了个饭馆，饭馆里有n个钩子，每个钩子放衣服得花ai元、若来的客人多了，钩子就不够挂衣服了，这个人就得支付给每个不能挂衣服的客人每人d元，问这个屌丝最大收益（可能为负值。）
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<ctype.h>
#include<algorithm>
#include<stack>
#include<queue>
#include<set>
#include<math.h>
#include<vector>
#include<map>
#include<deque>
#include<list>
using namespace std;
int main()
{
int n,d,m;
int a[105];
int w=0;
while(scanf("%d%d",&n,&d)!=EOF)
{
int sum=0;
for(int i=0; i<n; i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
scanf("%d",&m);
sort(a,a+n);
if(m>n)
w=sum-(m-n)*d;
else
{
for(int j=0; j<m; j++)
{
w+=a[j];
}
}
printf("%d\n",w);
}
return 0;
}