Problem:
Score : 200 points

Problem Statement

In some other world, today is the day before Christmas Eve.

Mr. Takaha is buying N items at a department store. The regular price of the i-th item (1 \leq i \leq N) is p_i yen (the currency of Japan).

He has a discount coupon, and can buy one item with the highest price for half the regular price. The remaining N-1 items cost their regular prices. What is the total amount he will pay?

Constraints

2 \leq N \leq 10

100 \leq p_i \leq 10000

p_i is an even number.

Input

Input is given from Standard Input in the following format:

N
p_1
p_2
:
p_N

Output

Print the total amount Mr. Takaha will pay.

Sample Input 1

3
4980
7980
6980

Sample Output 1

15950

The 7980-yen item gets the discount and the total is 4980 + 7980 / 2 + 6980 = 15950 yen.

Note that outputs such as 15950.0 will be judged as Wrong Answer.

Sample Input 2

4
4320
4320
4320
4320

Sample Output 2

15120

Only one of the four items gets the discount and the total is 4320 / 2 + 4320 + 4320 + 4320 = 15120 yen.

Solution in C (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
int main(void)
{
	int N,p[10000],max,T,i,j,k;

	scanf("%d\n",N);

	for(j=0;N>0;j++){
		scanf("%d\n",&p[i]);
                N--;}
	max=p[0];
	for (i=0;i<max;i++)
		if(max>p[i])
			max=p[i];

       for(k=0;N>0;k++){
	        T=T+p[i];
		N--;}
       T=T-1.5*max;

       printf("%d\n",T);
       return  0;

}