Bear and Five Cards

                                                                     time limit per test 2 seconds

                                                              memory limit per test 256 megabytes

                                                                            input standard input

                                                                          output standard output

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.

题解

题的大意是：小明、小刚和小红去买饮料，饮料的价格各不相同，一共挑了五种，三个人有可能会挑到相同种类的饮料，如果有2或者3个相同的，就不要这几个了（说明喝过都觉得好，既然喝过就算了吧，尝尝新的，还省钱不是），如果有四个以上相同的，就减去三个（喝过就别喝了。。其他的说不定是给别人带的，别人的意愿不好更改吧），其他的钱还想去游乐场，所以想知道买饮料最少用多少钱就够了

先排个序，好判断有没有相同的，有几个，如果有，记录下来，继续遍历，和之后可能能减的比较，取能减的最大值（相同的最多减三个）。这里从大到小或者从小到大排列都得从头遍历到尾，谁知道7+7和6+6+6谁大呢

# include <cstdio>
# include <algorithm>

using namespace std;
const int maxn = 1e5 + 10;

int main()
{
	int sum = 0;
	int a[10];
	for(int i=0;i<5;i++) 
	{
		scanf("%d",&a[i]);
		sum += a[i];
	} 
	
	sort(a,a+5);
	int maxm = 0, flag = 1, num;
	for(int i=0;i<5;i++)
	{
		if(a[i] == a[i+1]) 
		{
			flag++;
			num = a[i];
		}
		else if(flag>1) 
		{
			if(flag<4) maxm = max(maxm,num * flag);
			else maxm = max(maxm,num * 3);
			flag = 1;
		}
	}
	printf("%d\n",sum-maxm);
	return 0;
 }