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.io.*;
 
public class three {
 
	public static void main(String[] args) throws IOException, FileNotFoundException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		//BufferedReader in = new BufferedReader(new FileReader("three"));
 
		int t = Integer.parseInt(in.readLine());
		while (t --> 0) {
			String s = in.readLine();
			int n = s.length();
			int maxkeep = 0;
			
			for (int i=0; i<10; i++) {
				for (int j=0; j<10; j++) {
					char a = (char) (i + '0');
					char b = (char) (j + '0');
					boolean use_a = true;
					int cur=0;
					for (int k=0; k<n; k++) {
						if (use_a && s.charAt(k) == a) {
							cur++;
							use_a = false;
						}
						else if (!use_a && s.charAt(k) == b) {
							cur++;
							use_a = true;
						}
					}
					if (i!=j && cur%2==1) cur--;
					maxkeep = Math.max(maxkeep, cur);
				}
			}
			
			System.out.println(n - maxkeep);
			
		}
	}
}