CF845B.Luba And The Ticket

Luba has a ticket consisting of 6 digits. In one move she can choose digit in any position and replace it with arbitrary digit. She wants to know the minimum number of digits she needs to replace in order to make the ticket lucky.

The ticket is considered lucky if the sum of first three digits equals to the sum of last three digits.

卢巴有一张由 6 位数字组成的车票。在一次操作中，她可以选择任意位置上的数字，并将其替换为任意数字。她想知道，为了使这张车票变成“幸运车票”，她至少需要替换多少位数字。

当车票的前三位数字之和等于后三位数字之和时，该车票即被视为“幸运车票”。

You are given a string consisting of 6 characters (all characters are digits from 0 to 9) — this string denotes Luba's ticket. The ticket can start with the digit 0.

你将得到一个由 6 个字符组成的字符串（所有字符均为数字 0 到 9）——该字符串表示卢巴的车票。车票可能以数字 0 开头。

Print one number — the minimum possible number of digits Luba needs to replace to make the ticket lucky.

输出一个数字——Luba 需要替换的最少数字个数，以使这张票变为幸运票。

In the first example the ticket is already lucky, so the answer is 0.

In the second example Luba can replace 4 and 5 with zeroes, and the ticket will become lucky. It's easy to see that at least two replacements are required.

In the third example Luba can replace any zero with 3. It's easy to see that at least one replacement is required.

在第一个例子中，这张票已经是幸运的，因此答案为 0。

在第二个例子中，Luba 可以将 4 和 5 替换为 0，从而使这张票变为幸运的。显然，至少需要两次替换。

在第三个例子中，Luba 可以将任意一个 0 替换为 3。显然，至少需要一次替换。