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.*;

public class Solution
{
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
              int t = sc.nextInt();
              sc.nextLine();
              while(t-- > 0)
              {
                     String s = sc.nextLine();
                     
                     if(s.length() == 2)
                     System.out.println(0);
                     
                     else
                     {
                     int min = 100000000;
                     int freq[] = new int[10];
                     for(int i = 0 ; i < s.length() ; i++)
                     {
                            freq[s.charAt(i)-'0']++;
                     }
       
                     for(int i = 0 ; i < 10 ; i++)
                     {
                           
                            
                            if(freq[i] != 0)
                            {
                                    int x = 0;
                                  
                            x = (s.length()-freq[i]);
                            
                            min = Math.min(min , x);
                            if(freq[i] == s.length())
                            min = 0;
                            }
                           
                            
                           
                     }
                    
                     
                     for(int i = 0 ; i < 10 ; i++)
                     {
                            for(int j = 0 ; j < 10 ; j++)
                            {
                                   int curr = i;
                                   int len = 0;
                                 if(i != j)
                                 {
                                    for(int k = 0 ; k < s.length() ; k++)
                                    {
                                           if(s.charAt(k)-'0' == curr)
                                           {
                                                  len++;
                                                  curr = i+j-curr;
                                           }
                                    }
                                     if(len%2 == 1)
                                     len--;
                                     
                                     min = Math.min(min , s.length()-len);
                                 }
                            }
                     }
                     System.out.println(min);
                     }
              }
       }
}