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.*;
import java.util.*;
public class Ishu
{
	public static void main(String[] args)
	{
	Scanner scan=new Scanner(System.in);
	int t;
	t=scan.nextInt();
	while(t-->0)
		{
		int i;
		String str=scan.next();
		int n=str.length();
		int[] in=new int[10];
		int[][] index=new int[10][n];
		for(i=0;i<=9;++i)
			{
			in[i]=0;
			for(int j=0;j<n;++j)
				if((str.substring(j,j+1)).equals(String.valueOf(i)))
					index[i][in[i]++]=j;
			}
	/*	for(i=0;i<10;++i)
			{
			for(int j=0;j<in[i];++j)
				System.out.print(index[i][j]+" ");
			System.out.println();
			}*/
		int max=Integer.MIN_VALUE;
		for(i=0;i<=9;++i)
			{
			for(int j=0;j<=9;++j)
				{
				int sum=0;
				int ctr=0;
				for(int k=0;k<n;++k)
					{
					if(ctr%2==0)
						{
						if((str.substring(k,k+1)).equals(String.valueOf(i)))
							{
							++sum;
							++ctr;
							}
						}
					else
						{
						if((str.substring(k,k+1)).equals(String.valueOf(j)))
							{
							++sum;
							++ctr;
							}
						}
					if(sum%2==0)
						if(sum>max)
							max=sum;
					}
				}
			}
		for(i=0;i<10;++i)
			if(in[i]>max)
				max=in[i];
		System.out.println((n-max));
		}	
	}
}
