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 {
    static Scanner sc;
    static int l;
    static String S;
    static int[] num;
    static int maxL;
    public static void main(String[] args){
        sc = new Scanner(System.in);
        int n = sc.nextInt();
        for (int i=0;i<n;i++){
            Solve();
        }
    }
    private static void Solve(){
        S = sc.next();
        l = S.length();
        maxL = 2;
        num = new int[10];
        for (int i=0;i<l;i++){
            num[Integer.parseInt(String.valueOf(S.charAt(i)))]++;
        }
        for (int i=0;i<10;i++){
            maxL = Math.max(num[i],maxL);
        }
        for (int i=48;i<58;i++){
            for (int j=48;j<58;j++){
                maxL = Math.max(maxL,getFL((char) i,(char) j));
            }
        }
        System.out.println(l-maxL);

    }
    private static int getFL(char a,char b){
        boolean flag = true;
        int length = 0;
        for (int i=0;i<l;i++){
            if (flag){
                if (S.charAt(i)==a){
                    flag=false;
                    length++;
                }
            }else {
                if (S.charAt(i)==b){
                    flag=true;
                    length++;
                }
            }
        }
        if (length%2==0){
            return length;
        }else {
            return length-1;
        }
    }
}

/****
 *
 *  |s| == 2,  n-2
 *  |s| == ,  n-max
 *  |s|
 *
 *
 *   12121212
 *
 *
 *
 *
 * */