Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ACM-ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

A. Bear and Five Cards

time limit per test

2 secondsmemory limit per test

256 megabytesinput

standard inputoutput

standard outputA 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 *t*_{1}, *t*_{2}, *t*_{3}, *t*_{4} and *t*_{5} (1 ≤ *t*_{i} ≤ 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.

Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Sep/25/2018 05:16:41 (d3).

Desktop version, switch to mobile version.

User lists

Name |
---|