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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public final class C {

    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int t = Integer.parseInt(in.nextLine());
        for (int x = 0; x < t; x++) {
            final String str = in.nextLine();
            int res = 0;
            for (int a = 0; a <= 9; a++) {
                for (int b = 0; b <= 9; b++) {
                    int aCount = 0, bCount = 0;
                    for (char c : str.toCharArray()) {
                        final int d = c - '0';
                        if (aCount <= bCount) {
                            if (d == a) { aCount++; }
                        } else {
                            if (d == b) { bCount++; }
                        }
                    }
                    res = Math.max(res, a == b ? aCount + bCount : 2 * bCount);
                }
            }
            System.out.println(str.length() - res);
        }
    }
}
