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.
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner input = new Scanner(System.in);
		int test = input.nextInt();
		input.nextLine();
		while(test-->0){
		    char c[] = input.nextLine().toCharArray();
		    int n = c.length;
		    int count[] = new int[10];
		    int mx =0;
		    for(int i =0;i<c.length;i++){count[c[i]-'0']++;mx = Math.max(mx,count[c[i]-'0']);}
		    int res =Integer.MAX_VALUE;
		    res = Math.min(res,n-mx);
		    for(char i ='0';i<='9';i++){
		        for(char j = '0';j<='9';j++){
		            int countt =0;
		            boolean ok = true;
		            for(char cc:c){
		                if(cc == i && ok){
		                    ok = false;
		                    countt++;
		                }else if(cc == j && !ok){
		                    ok = true;
		                    countt++;
		                }
		            }
		            int val = countt/2;
		            if(val == 0)continue;
		            res = Math.min(res,n-(val*2));
		        }
		    }
		    System.out.println(res);
		}
	}
}
