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 Round_92;

import java.io.*;
import java.math.*;
import java.util.*;

public class Problem_C {  
    public static void main(String[] args) {
        FastReader in = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        
        int numTrials = in.nextInt();
        while(numTrials --> 0) {
            StringBuffer str = new StringBuffer(in.next());
            
            int[] nums = new int[str.length()];
            TreeSet<Integer> digits = new TreeSet<Integer>();
            
            for(int i = 0; i < str.length(); i++) {
            	int val = Integer.parseInt(str.substring(i, i + 1));
            	nums[i] = val;
            	digits.add(val);
            }
            
            ArrayList<Integer> check = new ArrayList<Integer>();
            check.addAll(digits);
            
            int maxLen = 0;
            for(int i = 0; i < check.size(); i++) {
            	for(int j = 0; j < check.size(); j++) {
            		int num1 = check.get(i), num2 = check.get(j);
            		int length = 0;
            		
            		boolean found = false;
        			for(int q = 0; q < nums.length; q++) {
        				if(!found && nums[q] == num1) {
        					found = true;
        					length++;
        				}
        				else if(found && nums[q] == num2) {
        					found = false;
        					length++;
        				}
        			}
        			
        			if(found == true && num1 != num2) {
        				length--;
        			}
            		
            		maxLen = Math.max(maxLen, length);
            	}
            }
            
            out.println(str.length() - maxLen);
        }
        
        out.flush();
    }
    
    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; 
        } 
    } 
}
