Description
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 a 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.
Input
The first line contains two integers n and d(1 ≤ n, d ≤ 100). The next line contains integers a, a, ..., a(1 ≤ a ≤ 100). The third line contains integer m(1 ≤ m ≤ 100).
Output
In a single line print a single integer — the answer to the problem.
Sample Input
2 1 2 1 2
3
2 1 2 1 10
-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.
题意：顾客挂衣服，每个称子可以挂一个衣服，每个钩子挂衣服所需要的价格不同，宾客会选便宜的 挂，钩子不够了房东要陪d元，现在有m个顾客，让求今晚房东收益；
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
class clothes{
public:
int n,d,m;
int a[];
clothes(int n,int d):n(n),d(d){
for(int i = ; i < n; i++){
cin >> a[i];
}
cin >> m;
}
int work(){
sort(a,a + n);
for(int i = n; i < m; i++)
a[i] = -d;
int ans = ;
for(int i = ; i < m; i++)
ans += a[i];
return ans;
}
};
int main(){
int n,d;
while(~scanf("%d%d", &n, &d)){
clothes _guest(n,d);
cout << _guest.work() << endl;
}
return ;
}