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.* ;
public class firstjava
{
	public static void main ( String[ ] args ) 
	{
		Scanner sc = new Scanner( System.in ) ;
		int T = sc.nextInt( ) ;
		sc.nextLine( ) ;
		while( T -- != 0 )
		{
			String s = new String ( sc.nextLine( ) ) ;
			int n = s.length( ) ;
			int answer = 0 ;
			for( int i = 0 ; i <= 9 ; i ++ )
			{
				for( int j = 0 ; j <= 9 ; j++ )
				{
					int ct = 0 ;
					int temp = 0 ;
					for( int k = 0 ; k < n ; k ++ )
					{
						if( i == j )
						{
							if( s.charAt( k ) == ( char )( 48 + i ) )
							{
								ct ++ ;
							} 	
							continue ;
						}
						if( s.charAt( k ) == ( char )( 48 + i ) && temp == 0 )
						{
							temp = 1 ;
						}
						else if( s.charAt( k ) == ( char )( 48 + j ) && temp == 1 )
						{
							ct += 2 ;
							temp = 0 ;
						}
					}
					answer = Math.max( answer , ct ) ;
				}
			}
			System.out.println( n - answer ) ;
		}
	}
}