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.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        
        Scanner scan=new Scanner(System.in);
        int t=scan.nextInt();
        while(t-->0)
        {
            String s=scan.next();
            int[] a=new int[10];
            int ans=0;
            int max=0;
            for(int i=0;i<s.length();i++)
            {
                a[s.charAt(i)-'0']++;
                max=Math.max(a[s.charAt(i)-'0'],max);
            }
            for(int i=0;i<10;i++)
            {
                for(int j=0;j<10;j++)
                {
                    int temp=0;
                    if(i==j)
                        continue;
                    boolean f1=true;
                    boolean f2=false;
                    
            for(int k=0;k<s.length();k++)
            {
                 if(s.charAt(k)-'0'==i&&f1)
                 {
                     temp++;
                     f2=true;
                     f1=false;
                 }else  if(s.charAt(k)-'0'==j&&f2)
                 {
                     temp++;
                     f1=true;
                     f2=false;
                 }
            }
                    if(temp%2!=0)
                    {
                        temp--;
                    }
                   // System.out.println("temp "+temp);
                    max=Math.max(temp,max);
                    
                }
            }
            System.out.println(s.length()-max);
        }
    }
}