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.*;
import java.util.*;

public class C {
    public static void main(String args[]) throws Exception {
        BufferedReader bu = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(bu.readLine());
        StringBuilder sb = new StringBuilder();
        while (t-- > 0) {
            String S = bu.readLine();
            int ans = 0;
            for (char a = '0'; a <= '9'; a++) {
                for (char b = '0'; b <= '9'; b++) {
                    char x = a, y = b;
                    int cnt = 0;
                    for (char c : S.toCharArray()) {
                        if (c == x) {
                            cnt++;
                            x = (char) (x^y^(y=x));
                        }
                    }
                    if (a == b) ans = Math.max(ans, cnt);
                    else ans = Math.max(ans, cnt&~1);
                }
            }
            sb.append(S.length()-ans+"\n");
        }
        System.out.print(sb);
    }
}
