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 goodString(string s) {
  int n = s.size();
  int global_max = 0;
  for (int i = 0; i <= 9; i++) {
    for (int j = 0; j <= 9; j++) {
      int local_max = 0;
      int _i = i;
      int _j = j;
      for (char c : s) {
        if (_i == c - '0') {
          local_max++;
          swap(_i, _j);
        }
      }
      if (i != j && local_max % 2 != 0) local_max--;
      global_max = max(global_max, local_max);
    }
  }
  return s.size() - global_max;
}
int main() {
  {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  };
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    cout << goodString(s) << endl;
  }
}
