Let's call left cyclic shift of some string t_1 t_2 t_3 ... t_{n - 1} t_n as string t_2 t_3 ... t_{n - 1} t_n t_1.

Analogically, let's call right cyclic shift of string t as string t_n t_1 t_2 t_3 ... t_{n - 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 ≤ 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 ⋅ 10^5). Each character s_i is digit 0–9.

It's guaranteed that the total length of strings doesn't exceed 2 ⋅ 10^5.

Output

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

Example

Input


3
95831
100120013
252525252525


Output


3
5
0

Note

In the first test case, you can erase any 3 characters, for example, the 1-st, the 3-rd, and the 4-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.
//package Codeforces;

import java.util.*;

public class GoodString {
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        
        while (t-->0)
        {
            String s = sc.next();
            
            long max = 0;
            for(int i=0;i<9;i++)
            {
                char c1 = (char)(i+48);
                for(int j=i+1;j<=9;j++)
                {
                    char c2 = (char)(j+48);
                    char last = '\u0000';
                    long count = 0;
                    for(int ind=0;ind<s.length();ind++)
                    {
                        char ch = s.charAt(ind);
                        if(ch==c1 || ch==c2)
                        {
                            if(last=='\u0000')
                            {
                                last = ch;
                                count++;
                            }
                            else if(last!=ch)
                            {
                                last = ch;
                                count++;
                            }
                        }
                    }
                    if(count%2==0)
                        max = Math.max(max, count);
                    else
                        max = Math.max(max, count-1);
                }
            }
            
            for(int i=0;i<=9;i++)
            {
                long count = 0;
                char c = (char)(i+48);
                for(int ind=0;ind<s.length();ind++)
                {
                    char ch = s.charAt(ind);
                    if(ch==c)
                    {
                        count++;
                    }
                }
                max = Math.max(max, count);
            }
            
            System.out.println(s.length()-max);
        }
    }
}
