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;
using ll = long long;
using pr = pair<int, int>;
int mx = std::numeric_limits<int>::max();
int main() {
  int t = 1;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    set<char> s_set;
    int min_del = mx;
    for (auto c : s) s_set.insert(c);
    for (auto a : s_set) {
      for (auto b : s_set) {
        int toDel = 0;
        if (a == b) {
          for (auto cc : s)
            if (cc != a) toDel++;
        } else {
          char curc = 'n';
          char first_c = 'n';
          int count = 0;
          for (int i = 0; i < int(s.size()); ++i) {
            if (s[i] == a or s[i] == b) {
              if (curc == 'n') {
                curc = s[i];
                first_c = curc;
                count++;
              } else if (curc == s[i])
                toDel++;
              else {
                curc = s[i];
                count++;
              }
            } else {
              toDel++;
            }
          }
          if (first_c == curc and count > 1) toDel++;
        }
        min_del = min(toDel, min_del);
      }
    }
    cout << min_del << endl;
  }
  return 0;
}
