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.Scanner;

public class C {
    public static void main(String[] args) {

        Scanner scn = new Scanner(System.in);
        int cases = scn.nextInt();
        for (int p = 0; p < cases; p++) {
            String nr = scn.next();
            int[] digits = new int[nr.length()];
            int maxRep = 0;
            for (int i = 0; i < nr.length(); i++) {
                int num = Integer.parseInt(nr.charAt(i) + "");
                digits[i] = num;
            }
            for (int i = 0; i < 10; i++) {
                for (int j = 0; j < 10; j++) {
                    int rep = repeatedPair(i, j, digits);
                    if (rep > maxRep) {
                        maxRep = rep;
                    }
                }
            }
            System.out.println(nr.length() - maxRep);
        }


    }

    public static int repeatedPair(int a, int b, int[] arr) {
        int ans = 0;
        if (a == b) {
            for (int i = 0; i < arr.length; i++) {
                if (arr[i] == a) {
                    ans++;

                }
            }
        } else {
            boolean turnA = true;
            for (int i = 0; i < arr.length; i++) {
                if (turnA && arr[i] == a) {
                    turnA = false;

                } else if (!turnA && arr[i] == b) {
                    turnA = true;
                    ans++;

                }
            }
            ans *= 2;
        }

        return ans;
    }
}
