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 HelloWorld{

    public static void main(String []args){
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        for(int i=0; i<t; i++){
            int[] arr = new int[10];
            String s = sc.next();
            int n = s.length();
            for(int j=0; j<n; j++){
                int rd = Character.getNumericValue(s.charAt(j));
                arr[rd]++;
            }
            int mx = 0, mx2 = 1;
            if(arr[1]>arr[0]){
                mx = 1;
                mx2 = 0;
            }
            for(int j=1; j<10; j++){
                if(arr[j]>arr[mx]){
                    mx2 = mx;
                    mx = j;
                } else if (arr[j]<=arr[mx]&&arr[j]>=arr[mx2]){
                    mx2 = j;
                }
            }
            int ans = arr[mx];
            for(int p=0; p<10; p++){
                for(int q=0; q<10; q++){
                    if(p==q) continue;
                    int a = arr[p], b = arr[q];
                    int vl = 0;
                    boolean fnd = true;
                    for(int ln=0; ln<n; ln++){
                        int ws = Character.getNumericValue(s.charAt(ln));
                        if(ws==p){
                            a--;
                            if(fnd){
                                vl++;
                                fnd = false;
                            }
                        }
                        if(ws==q){
                            b--;
                            if(!fnd){
                                vl++;
                                fnd = true;
                            }
                        }
                    }
                    if(vl%2==1){
                        ans = Math.max(ans,vl-1);
                    } else {
                        ans = Math.max(ans,vl);
                    }
                }
            }
            System.out.println(n-ans);
        }
    }
}