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 Solution {

    private static final String NO = "NO";
    private static final String YES = "YES";
    public static BufferedWriter bw;
    public static FR sc;
    private static final int MOD = 1000000007;

    public static void main(String[] args) throws IOException {
        sc = new FR();
        bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int t = 1;
        t = sc.i();
        while (t > 0) {
            t--;
            String s = sc.s();
            int ans = solve(s);
            bw.write(ans + "\n");
        }
        bw.close();

    }

    private static int solve(String s) {
        int len = s.length();
        if (len == 1)
            return 0;
        int ans = 0;
        for (int x = 0; x < 10; x++) {
            for (int y = 0; y < 10; y++) {
                int hey = getAns(s, len, x, y);
                if (hey > ans) ans = hey;
            }
        }
        return len - ans;
    }

    private static int getAns(String s, int len, int x, int y) {
        int ans = 0;
        boolean flag = true;
        char first = (char) ('0' + x);
        char second = (char) ('0' + y);

        for (int i = 0; i < len; i++) {
            if (s.charAt(i) == first && flag) {
                ans++;
                flag = false;
            } else if (s.charAt(i) == second && !flag) {
                ans++;
                flag = true;
            }
        }
        if (x != y)
            ans -= (ans % 2);
        return ans;
    }


}

class FR {
    BufferedReader br;
    StringTokenizer st;

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

    public String s() {
        if (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {

            }
        }
        return st.nextToken();
    }

    public int i() {
        return Integer.parseInt(s());
    }

    public Integer[] ia(int n) {
        Integer[] a = new Integer[n];
        for (int i = 0; i < n; i++) {
            a[i] = i();
        }
        return a;
    }

    public long[] la(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = l();
        }
        return a;
    }

    public String[] sa(int n) {
        String[] a = new String[n];
        for (int i = 0; i < n; i++) {
            a[i] = s();
        }
        return a;
    }

    public long l() {
        return Long.parseLong(s());
    }
}
