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.
/******************************************************************************

                            Online Java Compiler.
                Code, Compile, Run and Debug java program online.
Write your code in this editor and press "Run" button to execute it.

*******************************************************************************/
import java.util.*;
public class Main
{
    public static int solve(String s, int x, int y){
        int res=0;
        for(char c:s.toCharArray()){
            if((c-'0')==x){
                res++;
                int temp=x;
                x=y;
                y=temp;
                
            }    
        }
        if(x!=y && res%2==1)
            res--;
            return res;
    }
	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();
		    int ans=0;
		    for(int i=0;i<10;i++){
		        for(int j=0;j<10;j++){
		            ans = Math.max(ans,solve(s,i,j));
		        }
		    }
		    System.out.println(s.length()-ans);
		    
		}
	}
}
