CF1913B Swap and Delete

Description

You are given a binary string $ s $ (a string consisting only of 0-s and 1-s). You can perform two types of operations on $ s $ : 1. delete one character from $ s $ . This operation costs $ 1 $ coin; 2. swap any pair of characters in $ s $ . This operation is free (costs $ 0 $ coins). You can perform these operations any number of times and in any order. Let's name a string you've got after performing operations above as $ t $ . The string $ t $ is good if for each $ i $ from $ 1 $ to $ |t| $ $ t_i \neq s_i $ ( $ |t| $ is the length of the string $ t $ ). The empty string is always good. Note that you are comparing the resulting string $ t $ with the initial string $ s $ . What is the minimum total cost to make the string $ t $ good?

Input Format

The first line contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases. Then $ t $ test cases follow. The only line of each test case contains a binary string $ s $ ( $ 1 \le |s| \le 2 \cdot 10^5 $ ; $ s_i \in \{ $ 0, 1 $ \} $ ) — the initial string, consisting of characters 0 and/or 1. Additional constraint on the input: the total length of all strings $ s $ doesn't exceed $ 2 \cdot 10^5 $ .

Output Format

For each test case, print one integer — the minimum total cost to make string $ t $ good.

Explanation/Hint

In the first test case, you have to delete a character from $ s $ to get the empty string $ t $ . Only then $ t $ becomes good. One deletion costs $ 1 $ coin. In the second test case, you can, for example, delete the second character from $ s $ to get the string 01, and then swap the first and second characters to get the string $ t $ $ = $ 10. String $ t $ is good, since $ t_1 \neq s_1 $ and $ t_2 \neq s_2 $ . The total cost is $ 1 $ coin. In the third test case, you can, for example, swap $ s_1 $ with $ s_2 $ , swap $ s_3 $ with $ s_4 $ , swap $ s_5 $ with $ s_7 $ , $ s_6 $ with $ s_8 $ and $ s_9 $ with $ s_{10} $ . You'll get $ t $ $ = $ 1010001110. All swap operations are free, so the total cost is $ 0 $ .