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

public class C {
    public static void main(String[] args){
        FastReader in = new FastReader();
        int t = in.nextInt();
        while(t-- > 0){
            String s = in.nextLine();
            char[] ch = s.toCharArray();
            // If the string has odd length, then all characters should be equal to each other; otherwise, all characters on odd positions should be equal, and all characters on even positions should be equal.
            int[] freq = new int[10];
            for(int i = 0; i < ch.length; i++){
                freq[ch[i] - '0']++;
            }
            Arrays.sort(freq);
            int maxLen = freq[freq.length - 1];
            for(int first = 0; first < 10; first++){
                for(int second = 0; second < 10; second++){
                    int len = 0;
                    for(int k = 0; k < ch.length; k++){
                        if(len % 2 == 0){
                            if(ch[k] - '0' == first){
                                len++;
                            }
                        } else {
                            if(ch[k] - '0' == second){
                                len++;
                            }
                        }

                    }
                    len -= len % 2; //we want even length
                    maxLen = Math.max(maxLen, len);
                }
            }
            System.out.println(ch.length - maxLen);
        }

    }


    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}
