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.
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int max_count = 0;
    for (char ch = '0'; ch <= '9'; ch++) {
      int c1 = 0;
      for (char ch1 = '0'; ch1 <= '9'; ch1++) {
        int f = -1, c2 = 0;
        for (int i = 0; i < s.size(); i++) {
          if (s[i] == ch && f == -1) {
            c2++;
            f = 1;
          } else if (s[i] == ch1 && f == 1) {
            c2++;
            f = -1;
          }
        }
        if (c2 % 2 != 0) c2--;
        max_count = max(max_count, c2);
      }
      for (int i = 0; i < s.size(); i++) {
        if (s[i] == ch) c1++;
      }
      max_count = max(c1, max_count);
    }
    int res = s.size() - max_count;
    cout << res << endl;
  }
  return 0;
}
