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.*;
import java.io.*;

public class Solution {
   
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        StringBuilder res = new StringBuilder();
        int t = sc.nextInt();
        while (t-- > 0) {
            String str = sc.next();
            int[] count = new int[10];
            for (int i = 0; i < str.length(); i++) {
                count[str.charAt(i) - '0']++;
            }
            int n = str.length();
            //Make all equal
            int ans = Integer.MAX_VALUE;
            for (int i = 0; i < 10; i++) {
                ans = Math.min(ans, n - count[i]);
            }
            for (int i = 0; i < 10; i++) {
                for (int j = 0; j < 10; j++) {
                    int c = 0;
                    int len = 0;
                    int flag = 0;
                    for (int k = 0; k < n; k++) {
                        int curr = str.charAt(k) - '0';
                        if (flag == 0 && curr == i) {
                            len++;
                            flag ^= 1;
                        } else if (flag == 1 && curr == j) {
                            len++;
                            flag ^= 1;
                        } else {
                            c++;
                        }
                    }
                    if (len % 2 == 1 && len != 1)
                        c++;
                    ans = Math.min(c, ans);
                }
            }
            pw.println(ans);
 
        }
        pw.close();
    
    }
 
 
}