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 CodeForces1389C{
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int t = input.nextInt();
		for(int i = 0;i<t;i++){
			String s = input.next();
			int[] a = new int[10];
			int[][] aldFreq = new int[10][10];
			for(char c: s.toCharArray()){
				int n = c - '0';
				a[n]++;
				for(int j = 0;j<10;j++){
					if(j == n){
						continue;
					}
					aldFreq[j][n] = aldFreq[n][j] + 1;
				}
			}
			int max = 0;
			for(int[] row: aldFreq){
				for(int j: row){
					max = Math.max(max,j);
				}
			}
			if(max%2 == 1){
				max--;
			}

			for(int j: a){
				max = Math.max(max,j);
			}

			System.out.println(s.length() - max);
		}
	}
}