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 Educational92;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class ProblemC {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder print=new StringBuilder();
        StringTokenizer st;
        int test=Integer.parseInt(br.readLine());
        while(test--!=0){
            char c[]=br.readLine().toCharArray();
            print.append(solve(c)+"\n");
        }
        System.out.print(print.toString());
    }
    public static int solve(char c[]){
        int n=c.length;
        int ans=n;
        int count[]=new int[10];
        for(int i=0;i<n;i++){
            count[c[i]-48]++;
        }
        for(int i=0;i<10;i++){
            ans=Math.min(ans,n-count[i]);
        }
//        System.out.println(ans);
        for(int i=48;i<=57;i++){
            for(int j=48;j<=57;j++){
                if(i==j){
                    continue;
                }
                boolean turn=true;
                int rem=0;
                for(int k=0;k<n;k++){
                    if(turn&&c[k]==i){
                        turn=!turn;
                        rem++;
                    }
                    if(!turn&&c[k]==j){
                        turn=!turn;
                        rem++;
                    }
                }
//                System.out.println(rem+" "+i+" "+j);
                if(rem%2==0)
                ans=Math.min(ans,n-rem);
            }
        }
        return ans;
    }
}
