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 C {
	public static void main(String[] args) {
		Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

		int T = sc.nextInt();
		while (T-- > 0) {
			char[] s = sc.next().toCharArray();
			
			int ans = 0;
			for (char a = '0'; a <= '9'; a++) {
				int count = 0;
				for (char d : s) {
					if (a == d) {
						count++;
					}
				}
				ans = Math.max(ans, count);
				for (char b = (char) (a + 1); b <= '9'; b++) {
					ans = Math.max(ans, check(s, b, a));
					ans = Math.max(ans, check(s, a, b));
				}
			}
			
			out.println(s.length - ans);
		}

		sc.close();
		out.close();
	}
	
	static int check(char[] s, char a, char b) {
		char x = a;
		int count = 0;
		for (char c : s) {
			if (c == x) {
				x = x == a ? b : a;
				count++;
			}
		}
		return count / 2 * 2;
	}
}
