C. Good String Educational Codeforces Round 92 (Rated for Div. 2)

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文章讨论了如何通过删除数字字符，使得给定的数字字符串在左右移位后结果相等。提供了一个编程解题思路，涉及字符计数和条件判断。

Let's call left cyclic shift of some string t1t2t3…tn−1tn�1�2�3…��−1�� as string t2t3…tn−1tnt1�2�3…��−1��1.

Analogically, let's call right cyclic shift of string t� as string tnt1t2t3…tn−1��1�2�3…��−1.

Let's say string t� is good if its left cyclic shift is equal to its right cyclic shift.

You are given string s� which consists of digits 0–9.

What is the minimum number of characters you need to erase from s� to make it good?

Input

The first line contains single integer t� (1≤t≤10001≤�≤1000) — the number of test cases.

Next t� lines contains test cases — one per line. The first and only line of each test case contains string s� (2≤|s|≤2⋅1052≤|�|≤2⋅105). Each character si�� is digit 0–9.

It's guaranteed that the total length of strings doesn't exceed 2⋅1052⋅105.

Output

For each test case, print the minimum number of characters you need to erase from s� to make it good.

Example

input

Copy

3
95831
100120013
252525252525

output

Copy

3
5
0

Note

In the first test case, you can erase any 33 characters, for example, the 11-st, the 33-rd, and the 44-th. You'll get string 51 and it is good.

In the second test case, we can erase all characters except 0: the remaining string is 0000 and it's good.

In the third test case, the given string s� is already good.

题目大意：

给一个数字字符串，在删掉其中一些数字字符后，使得其左移x位和右移x位的结果相等。

问最少要删除多少数字？

思路：

能使数字字符串要使得其左移x位和右移x位的结果相等的字符串有两种：

1、只有一种数字字符。

2、只有两种数字字符且数量相等，还需要错位排列。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"

void solve(){
	string s;
	cin >> s;
	ll sum=0,len=s.size();
	for(char x = '0' ; x <= '9' ; x ++)
		for(char y = '0' ; y <= '9' ; y ++){
			ll cnt = 0;
			char c=x;
			for(ll i = 0 ; i < len ; i ++)
				if(s[i] == c)
					cnt++ ,c = c == x ? y : x;
			sum = max(sum,cnt-(x != y && cnt&1) );
		}
	cout << len-sum << endl;
	return;
}

int main(){
	ll t=1;cin >> t;
	while(t--)solve();
	return 0;
}