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.*;
import java.io.*;

public class C {

    public static void main(String[] args) throws Exception {
        FastScanner in = new FastScanner();
        int T = in.nextInt();
        for (int R = 0; R < T; R++) {
            char[] ac = in.next().toCharArray();
            int n = ac.length;
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ac[i] - '0';
            }
            int max1 = 0;
            int[] ar = new int[10];
            for (int x : a) {
                ar[x]++;
            }
            for (int x : ar) {
                max1 = Math.max(max1, x);
            }
            int min = Integer.MAX_VALUE;
            for (int i = 0; i < 10; i++) {
                for (int j = 0; j < 10; j++) {
                    if (i == j)
                        continue;
                    boolean first = true;
                    int sum = 0;
                    for (int x = 0; x < n; x++) {
                        if (a[x] != i && a[x] != j) {
                            sum++;
                            continue;
                        } else if (first && a[x] != i) {
                            sum++;
                            continue;
                        } else if (!first && a[x] != j) {
                            sum++;
                            continue;
                        } else {
                            first = !first;
                        }
                    }
                    if (!first) {
                        sum++;
                    }
                    min = Math.min(min, sum);
                }
            }
            System.out.println(Math.min(n - max1, min));
        }
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}
