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 GoodString {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T = in.nextInt();
        for (int t = 0; t < T; t++) {
            String s = in.next();
            int[] freq = new int[10];
            int[][] altFreq = new int[10][10];
            for (char c : s.toCharArray()) {
                int n = c - '0';
                freq[n]++;
                for (int i = 0; i < 10; i++) {
                    if (i == n) {
                        continue;
                    }
                    altFreq[i][n] = altFreq[n][i] + 1;
                }
            }
            int max = 0;
            for (int[] row : altFreq) {
                for (int i : row) {
                    max = Math.max(max, i);
                }
            }
            if (max % 2 == 1) {
                max--;
            }
            for (int i : freq) {
                max = Math.max(max, i);
            }
            System.out.println(s.length() - max);
        }
    }
}