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.
//package com.company;

import java.util.Scanner;

public class CF_1389_C {
    public static void main(String args[]) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        while(t-->0) {
            String s=sc.next();
//            int ll[][] = new int[10][10];
//            int ch[]=new int[10];
//            for (int i = 0; i < s.length(); ++i) {
//                char c=s.charAt(i);
//                for(int j=0;j<10;++j) {
//                    ll[j][c-'0']+=ch[j];
//                }
//                ch[i-]
//            }
            int max=0;
            int ch[]=new int[10];
            for(int i=0;i<s.length();++i)
                ch[s.charAt(i)-'0']++;
            for(int i:ch)
                max=Math.max(i,max);
            for(int i=0;i<10;++i) {
                for(int j=0;j<10;++j) {
                    if(i==j)
                        continue;
                    int count=0;
                    int num=0;
                    for(int k=0;k<s.length();++k) {
                        if(count==0) {
                            if(s.charAt(k)-'0'==i) {
                               num++;
                               count=1;
                            }
                        } else {
                            if(s.charAt(k)-'0'==j) {
                                num++;
                                count=0;
                            }
                        }
                    }
                    if(num%2==1)
                        num--;
                    max=Math.max(num,max);
                }
            }
            System.out.println(s.length()-max);
        }
    }
}
