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.
import java.io.*;
import java.util.*;

public class sort {
    public static int gcd(int a,int b)
    {
        if(b==0)
         return a;
        else
         return gcd(b,a%b); 
    }
    public static void main(String[] args) throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(br.readLine());
        StringBuffer sb=new StringBuffer();
        while(t>0)
        {
            t--;
            
            String s=br.readLine();
            int n=s.length();
            int[] a=new int[10];int max=0;
            for(int i=0;i<n;i++)
            { a[s.charAt(i)-'0']++;
              max=Math.max(max,a[s.charAt(i)-'0']);
            }
            for(int i=0;i<=9;i++)
            {
                for(int j=0;j<=9;j++)
                {
                    if(i==j)
                     continue;
                    char tos=(char)(i+48); int l=0;
                    for(int k=0;k<n;k++)
                    {
                       if(s.charAt(k)==tos)
                       {
                           l++;
                           if(tos==(char)(i+48))
                            tos=(char)(j+48);
                           else
                            tos=(char)(i+48); 
                       }
                    }
                    if(l%2==1)
                     l--;
                    max=Math.max(l,max); 
                }
            }
            sb.append(n-max);    
            if(t!=0)
             sb.append(System.lineSeparator());  
        }
        System.out.println(sb);
    }
}