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.util.Scanner;
import java.util.HashMap;
import java.util.HashSet;
import java.util.ArrayList;
public class Main{
    public static void main(String []args)
    {
      Scanner scn = new Scanner(System.in);
      int t = scn.nextInt();
      while(t-->0)
      {
          String s = scn.next();
       
          int maximum=0;
          for(int k=0;k<10;k++)
          {
              for(int j=0;j<10;j++)
              {
            	   int cnt=0;
                  boolean flag=false;
                  char f = (char)(k+'0');
                  char sec = (char)(j+'0');
                  for(int i=0;i<s.length();i++)
                  {
                     if(!flag&&s.charAt(i)==f)
                     {
                         cnt++;
                         flag=true;
                     }else if(flag&&s.charAt(i)==sec)
                     {
                         cnt++;
                         flag=false;
                     }
                  }
                  if(cnt%2==1&&f!=sec)
                  {
                	  cnt--;
                  }
                  maximum = Math.max(cnt,maximum);
              }
          }
          System.out.println(s.length()-maximum);
      }
      
       
  }
  
  
 
}