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 _1389C {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		while (t-- > 0) {
			String s = sc.next();
			int max = 0;
			for (int i = 0; i < 10; i++) {
				int len = 0;
				for (int j = 0; j < s.length(); j++) {
					if (s.charAt(j)-'0' == i) {
						len++;
					}
				}
				max = Math.max(max, len);
			}
			for (int i = 0; i < 10; i++) {
				for (int j = 0; j < 10; j++) {
					if (i != j) {
						boolean first = true;
						int len = 0;
						for (int k = 0; k < s.length(); k++) {
							if (first && s.charAt(k)-'0' == i) {
								len++;
								first = !first;
							}
							else if (!first && s.charAt(k)-'0' == j) {
								len++;
								first = !first;
							}
						}
						len -= len%2;
						max = Math.max(max, len);
					}
				}
			}
			System.out.println(s.length()-max);
		}

	}

}
