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 Solution {

    public static void main(String[] args) {
        Scanner s= new Scanner(System.in);
        int t=s.nextInt();
        while(t-->0)
        {
            String st=s.next();
            
            int[] count=new int[10];
            for(int i=0;i<st.length();i++)
                count[st.charAt(i)-'0']++;
                
            int pair=-1;    
            for(int i=0;i<10;i++)
            {
                for(int j=i+1;j<10;j++)
                {
                    if(count[i]==0 || count[j]==0)
                        continue;

                    int a=0;
                    int b=0;
                    for(int k=0;k<st.length();k++)
                    {
                        int x=st.charAt(k)-'0';
                        if(x==i)
                            a++;
                        else if(x==j)
                        {
                            if(a!=0)
                            {
                                b++;
                                a=0;
                            }
                        }
                    }
                    pair=Math.max(pair,b);

                    a=0;
                    b=0;
                    for(int k=0;k<st.length();k++)
                    {
                        int x=st.charAt(k)-'0';
                        if(x==j)
                            a++;
                        else if(x==i)
                        {
                            if(a!=0)
                            {
                                b++;
                                a=0;
                            }
                        }
                    }
                    pair=Math.max(pair,b);
                }
            }

            int ans=Integer.MAX_VALUE;
            if(pair!=-1)
                ans=st.length()-(2*pair);
            for(int i=0;i<10;i++)
            {
                int a=st.length()-count[i];
                ans=Math.min(ans,a);
            }
            System.out.println(ans);
        }
    }
}