Codeforces Round #215 (Div. 2) -A. Sereja and Coat Rack

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本文介绍了一个关于餐厅收益计算的问题，讨论了如何在有限的衣帽架挂钩数量下，最大化收益的同时处理额外顾客的罚款情况。通过排序和条件判断实现了算法的有效解决。

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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_i 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 a₁, a₂, ..., a_n (1 ≤ a_i ≤ 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 test(s)

Input

2 1
2 1
2

Output

Input

2 1
2 1
10

Output

-5

Note

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;
}