I - Sereja and Coat Rack

I - Sereja and Coat Rack

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

Submit Status

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 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.

Input

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).

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.

题目分析：

没有什么难度，意思就是

2 1

2 1

10

一个小饭馆有2个衣架如果有客人来没有衣架了那么老板就要给客人1块钱补偿

第二行   是每个衣架使用的费用，

第三行就是  来的客人总数

注意：注意先来的客人肯定是选最便宜的衣架了，你懂的，如果是你你不选便宜的吗？

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
  int n,k,m;
  int a[1100];
  while(~scanf("%d%d",&n,&k))
  {
    for(int i=0;i<n;i++)
      scanf("%d",&a[i]);
    scanf("%d",&m);
    sort(a,a+n);
//    for(int i=1;i<=n;i++)
//      printf("%d ",a[i]);
//      printf("\n");
    int cost=0;
    if(m<=n)
    {
      for(int i=0;i<m;i++)
        cost+=a[i];
    }
    else
    {
      for(int i=0;i<n;i++)
        cost+=a[i];
      cost-=(m-n)*k; 
    }
    //if(m==0) cost=0;
    printf("%d\n",cost);
  }
  return 0;
}