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.util.*;

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

public class Template {
	
	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;
  }
  
}
	
	static void printMatrix(int mat[][])
	{
		for(int i=0;i<mat.length;i++)
		{
			for(int j=0;j<mat[0].length;j++)
			{
				System.out.print(mat[i][j]+" ");
			}
			System.out.println();
		}
	}
	
	static void printArray(int a[])
	{
		for(int i=0;i<a.length;i++)	
		{
			System.out.print(a[i]+" ");
		}
		System.out.println();
		
	}
	

//	    public static void main(String args[])
//	    {
//	    	FastReader sc = new FastReader();
//	    	//solve();
//	    	//Scanner sc = new Scanner(System.in)
//	       int testcases = sc.nextInt();  // nTest is the number of treasure hunts.
//	      
////	    	int testcases = 3;
//	        while(testcases-- > 0)
//	        {
//	        	String s = sc.next();
//	        	solve(s);
//	        	
//	        }
//	    
//	    }
	
	static int[][][]dp;
	static int []a;
	static int solve(int z, int last, int ind, int movesLeft)
	{
		if(dp[z][last][ind] != -1)
			return dp[z][last][ind];
		
		if(movesLeft == 0)
			return 0;
		
		int ans = 0;
		if(last == 0 && ind > 0 && z > 0)
		{
			ans = Math.max(ans, solve(z-1, 1, ind-1, movesLeft-1) + a[ind-1]);
		}
		
		ans = Math.max(ans, solve(z, 0, ind+1, movesLeft-1) + a[ind+1]);
		return  dp[z][last][ind] = ans;
	}

	public static void main(String[] args) {
	
		
		
		FastReader sc=new FastReader();
		//PrintWriter out = new PrintWriter(System.out);
		int T=sc.nextInt();

		outer:for (int tt=0; tt<T; tt++) {
			
			char a[] = sc.next().toCharArray();
			int map[] = new int[10];
			
			for(int i=0;i<a.length;i++)
			{
				map[a[i] - '0']++;
			}
			
			Arrays.sort(map);
			int ans = map[9];
			
			for(int first = 0;first<10;first++)
			{
				for(int second = 0;second < 10;second++)
				{
					int len = 0;
					for(char c:a)
					{
						if(len %2 == 0)
						{
							if(c -'0' == first) len++;
						}
						else
							if(c - '0' == second) len++;
					}
					len-=len%2;
					ans = Math.max(ans, len);
				}
			}
			System.out.println(a.length - ans);
		}
	
	}
}