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 C {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int t = input.nextInt();
        input.nextLine();
        for(int i = 0; i < t; i++) {
            String s = input.nextLine();
            int min = Integer.MAX_VALUE;
            min = Math.min(min, s.length() - 2);
            int[] nums = new int[10];
            for(int k = 0; k < s.length(); k++) {
                nums[Integer.parseInt(s.substring(k, k + 1))]++;
            }
            int temp = nums[0];
            for(int k = 1; k < 10; k++) {
                temp = Math.max(temp, nums[k]);
            }
            min = Math.min(min, s.length() - temp);
            int[][] times = new int[10][10];
            ArrayList<Integer> help = new ArrayList<>();
            for(int k = 0; k < s.length(); k++) {
                int temptemp = Integer.parseInt(s.substring(k, k + 1));
                if(help.contains(temptemp)) {
                    int pos = 0;
                    for(int h = 0; h < help.size(); h++) {
                        if(help.get(h) == temptemp) {
                            pos = h;
                            break;
                        }
                    }
                    for(int h = pos + 1; h < help.size(); h++) {
                        times[help.get(h)][temptemp]++;
                    }
                    help.remove(pos);
                    help.add(temptemp);
                } else {
                    for(int h = 0; h < help.size(); h++) {
                        times[help.get(h)][temptemp]++;
                    }
                    help.add(temptemp);
                }
            }
            int max = 0;
            for(int x = 0; x < 10; x++) {
                for(int y = 0; y < 10; y++) {
                    max = Math.max(times[x][y], max);
                }
            }
            min = Math.min(min, s.length() - (max * 2));
            System.out.println(min);
        }
    }
}
