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.io.*;
import java.util.*;

public class Test {

    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        int t = in.nextInt();
        String s;
        int ans;
        List<List<Integer>> l;
        while (t-- > 0) {
            l = new ArrayList<>();
            for (int i = 0; i < 10; i++) {
                l.add(new ArrayList<>());
            }
            s = in.next();
            ans = 0;
            for (int i = 0; i < s.length(); i++) {
                l.get(Integer.parseInt(s.charAt(i) + "")).add(i);
            }
            for (int i = 0; i < 10; i++) {
                if (l.get(i).size() > ans) {
                    ans = l.get(i).size();
                }
            }
            int ii, jj, tmp;
            for (int i = 0; i < 10; i++) {
                for (int j = i + 1; j < 10; j++) {
                    ii = 0;
                    jj = 0;
                    tmp = 0;
                    if (!l.get(i).isEmpty() && !l.get(j).isEmpty()) {
                        if (l.get(j).get(jj) < l.get(i).get(ii)) {
                            tmp = 1;
                            while (jj < l.get(j).size() && l.get(j).get(jj) < l.get(i).get(ii)) {
                                jj++;
                            }
                        }
                    }
                    while (ii < l.get(i).size() && jj < l.get(j).size()) {
                        tmp++;
                        while (ii < l.get(i).size() && l.get(i).get(ii) < l.get(j).get(jj)) {
                            ii++;
                        }
                        tmp++;
                        while (ii < l.get(i).size() && jj < l.get(j).size() && l.get(j).get(jj) < l.get(i).get(ii)) {
                            jj++;
                        }
                    }
                    if (ii < l.get(i).size()) {
                        tmp++;
                    }
                    if (tmp % 2 == 1) {
                        tmp--;
                    }
                    if (tmp > ans) {
                        ans = tmp;
                    }
                }
            }
            System.out.println(s.length()-ans);
        }
    }

}

class FastScanner {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer("");

    String next() {
        while (!st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int nextInt() {
        return Integer.parseInt(next());
    }

    int[] readArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        return a;
    }

    long nextLong() {
        return Long.parseLong(next());
    }
}
