A. Sereja and Coat Rack

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

input

2 1
2 1
2

output

3

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.

解题说明：此题其实就是一个简单的排序，找出n个数中最小的m个数求和，如果m大于n，则减去d*(m-n)即可。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include <cstring>
using namespace std;

int main()
{
	int i, k, n, m, d, a[110];
	memset(a, 0, sizeof(a));
	scanf("%d %d", &n, &d);
	for (i = 0; i < n; i++)
	{
		scanf("%d", &a[i]);
	}
	scanf("%d", &m);
	sort(a , a + n);
	k = 0;
	for (i = 0; i < m; i++)
	{
		if (a[i] != 0)
		{
			k += a[i];
		}
		else
		{
			k -= d;
		}
	}
	printf("%d\n", k);
	return 0;
}