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 GoodString {
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);

		int t = input.nextInt();
		input.nextLine();

		while (t > 0) {
			String s = input.nextLine();

			int n = s.length();

			int ans = 0;
			for (int i = 0; i < 10; i++) {
				for (int j = 0; j < 10; j++) {
					int x = i, y = j, cnt = 0;
				
					for (int k = 0; k < n; k++) {
						if ((s.charAt(k) - '0') == x) {
							cnt++;

							int tmp = x;
							x = y;
							y = tmp;
						}
					}

					if (i != j && cnt % 2 != 0) {
						cnt--;
					}

					ans = Math.max(ans, cnt);
				}
			}

			System.out.println(n - ans);

			t--;
		}

		input.close();
	}
}
