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 Main {

    public static void main(String[] args) {

        Scanner input = new Scanner(System.in);

        int len = input.nextInt();
        input.nextLine();
        for(int i = 0; i < len; i++) {
            String str = input.nextLine();
            answer(str);
        }

    }

    public static void answer(String str) {

        // can be all the same
        // will have to be in pairs of two if string is even

        // brute force pattern match to see what pairs are closest to actual string

        int minChange = Integer.MAX_VALUE;
        for(int i = 0; i < 10; i++) {
            for(int j = 0; j < 10; j++) {

                // length of new string
                int newLen = 0;

                //  current alternating numbers
                int startVal = i;
                int endVal = j;

                for(char c : str.toCharArray()) {

                    if(c - '0' == startVal) {
                        // when we move forward we swap which value we want to see first
                        int temp = startVal;
                        startVal = endVal;
                        endVal = temp;

                        newLen++;
                    }

                }

                // if string is odd and not the same number, shave off a number from the string
                if(i != j && newLen % 2 == 1) {
                    newLen--;
                }

                // string length - newlength is the amount of characters needed to remove
                minChange = Math.min(minChange, str.length() - newLen);

            }
        }

        System.out.println(minChange);

    }

}

 		  			 	  		  				 	 		 		 		