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: Wrong Answer)

#include<stdio.h>
int main(void){
	
	int N;
	scanf("%d",&N);
	
	int p[10];
	int i;
	for(i = 0;i < N;i++){
		scanf("%d",&p[i]);
	}
	int minnum,minp = p[0];
	
	for(i = 0;i < N;i++){
		if(p[i] <= minp){
			minp = p[i];
			minnum = i;
		}
	}
	
	int sum = 0;
	for(i=0;i<N;i++){
		if(i == minnum){
			sum = sum + p[i]/2;
		}else{
			sum = sum + p[i];
		}
	}
	
	printf("%d",sum);
	return 0;
}