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.io.PrintWriter;
import java.util.StringTokenizer;

public class TaskC {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;
        PrintWriter pw = new PrintWriter(System.out);
        
        try {
            int t = Integer.parseInt(br.readLine());
            while(t-->0) {
                char[] c = br.readLine().toCharArray();
                int n = c.length;
                int[] a = new int[10];
                for(int i=0 ; i<n ; i++) {
                    a[c[i] - '0']++;
                }

                int min = n-2;
                for(int i=0 ; i<10 ; i++) {
                    min = Math.min(min, n-a[i]);
                }

                for(char x = '0' ; x<='9' ; x++) {
                    for(char y='0' ; y<='9' ; y++) {
                        if(x != y) {
                            boolean b = true;
                            int str = 0;
                            for(int i=0 ; i<n ; i++) {
                                if(b) {
                                    if(c[i] == x) {
                                        str++;
                                        b = !b;
                                    }
                                }
                                else {
                                    if(c[i] == y) {
                                        str++;
                                        b = !b;
                                    }
                                }
                            }
                            if(str%2 == 1) str--;
                            min = Math.min(min, n-str);
                        }
                    }
                }

                pw.println(min);
            }
        }
        finally {
            pw.flush();
            pw.close();
        }
    }
}
