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.Scanner;

public class er92c {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int t=scan.nextInt();
		for(int tt=0;tt<t;tt++) {
			char[] a=scan.next().toCharArray();
			int n=a.length;
			
			int res=Integer.MAX_VALUE;
			
			for(char i='0';i<='9';i++) {
				for(char j='0';j<='9';j++) {
					char cur=i;
					int len=0;
					for(int x=0;x<n;x++) {
						if(a[x]==cur) {
							len++;
							if(cur==i) cur=j;
							else cur=i;
						}
					}
					if(i!=j) len-=len%2;
					res=Math.min(res,n-len);
				}
			}
			System.out.println(res);
		}
	}
}