Codeforces Round #356 (Div. 2)-A

A. Bear and Five Cards

题目链接：http://codeforces.com/contest/680/problem/A

A little bear Limak plays a game. He has five cards. There is one number written on each card. Each number is a positive integer.

Limak can discard (throw out) some cards. His goal is to minimize the sum of numbers written on remaining (not discarded) cards.

He is allowed to at most once discard two or three cards with the same number. Of course, he won't discard cards if it's impossible to choose two or three cards with the same number.

Given five numbers written on cards, cay you find the minimum sum of numbers on remaining cards?

Input

The only line of the input contains five integers t1, t2, t3, t4 and t5 (1 ≤ ti ≤ 100) — numbers written on cards.

Output

Print the minimum possible sum of numbers written on remaining cards.

Examples

input

7 3 7 3 20

output

26

input

7 9 3 1 8

output

28

input

10 10 10 10 10

output

20

Note

In the first sample, Limak has cards with numbers 7, 3, 7, 3 and 20. Limak can do one of the following.

• Do nothing and the sum would be 7 + 3 + 7 + 3 + 20 = 40.
• Remove two cards with a number 7. The remaining sum would be 3 + 3 + 20 = 26.
• Remove two cards with a number 3. The remaining sum would be 7 + 7 + 20 = 34.

You are asked to minimize the sum so the answer is 26.

In the second sample, it's impossible to find two or three cards with the same number. Hence, Limak does nothing and the sum is 7 + 9 + 1 + 3 + 8 = 28.

In the third sample, all cards have the same number. It's optimal to discard any three cards. The sum of two remaining numbers is 10 + 10 = 20.

题意：有5张卡，每张卡有一个数字。现在可以丢弃一些卡，丢弃的卡要满足：存在2-3张数字一样的卡。 问最小的数字总和是多少？

思路：只要5张卡，暴力就好，注意要存在2-3张数字一样的卡才能丢弃，而且最多丢弃1次，数量最多是3张

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
const int MAXN=+;
typedef long long int LL;
#define INF 0x3f3f3f3f
int T[MAXN],tot[];
int main()
{
    while(~scanf("%d%d%d%d%d",&T[],&T[],&T[],&T[],&T[])){
        int sum=;
        memset(tot,,sizeof(tot));
        for(int i=;i<;i++){
            sum+=T[i];
            tot[T[i]]++;
        }
        int maxval=;
        for(int i=;i<=;i++){
            maxval=max(maxval,i*min(,(tot[i]>=?tot[i]:)));
        }
        printf("%d\n",sum-maxval);
    }
    return ;
}