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.math.BigInteger;
import java.util.*;


public class C {

  void solve() throws IOException {
    int t = nextInt();
    while (t-->0) {
      String s = nextToken();
      int ans = s.length() - 1;
      int[] count = new int[10];
      for (int i = 0; i < s.length(); i++) count[s.charAt(i) - '0']++;
      for (int i = 0; i < 10; i++) {
        ans = Math.min(ans, s.length() - count[i]);
      }
      for (int x = 0; x <= 10; x++) {
        for (int y = 0; y <= 10; y++) {
          if (x == y) {
            continue;
          }
          int deleteCount  = 0;
          int cur = x;
          int next = y;
          for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) - '0' == cur) {
              int tmp = cur;
              cur = next;
              next = tmp;
            } else {
              deleteCount++;
            }
          }
          if (cur == y) deleteCount++;
          ans = Math.min(ans, deleteCount);
        }
      }
      out.println(ans);
    }
  }

  public static void main(String[] args) throws IOException {
    new C().run();
  }

  void run() throws IOException {
    reader = new BufferedReader(new InputStreamReader(System.in));
//		reader = new BufferedReader(new FileReader("input.txt"));
    tokenizer = null;
    out = new PrintWriter(new OutputStreamWriter(System.out));
//		out = new PrintWriter(new FileWriter("output.txt"));
    solve();
    reader.close();
    out.flush();

  }

  BufferedReader reader;
  StringTokenizer tokenizer;
  PrintWriter out;

  int nextInt() throws IOException {
    return Integer.parseInt(nextToken());
  }

  long nextLong() throws IOException {
    return Long.parseLong(nextToken());
  }

  double nextDouble() throws IOException {
    return Double.parseDouble(nextToken());
  }

  String nextToken() throws IOException {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine());
    }
    return tokenizer.nextToken();
  }
}
