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(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    map<char, long long> mp;
    for (char c : s) mp[c]++;
    long long ans = 1e18;
    for (char i = '0'; i <= '9'; i++) {
      for (char j = '0'; j <= '9'; j++) {
        long long l = 0;
        string str = "";
        for (int k = 0; k < s.size(); k++) {
          if (l & 1) {
            if (s[k] == i) l++, str += s[k];
          } else {
            if (s[k] == j) l++, str += s[k];
          }
        }
        if (str.size() && i != j)
          if (str[str.size() - 1] == str[0]) str.pop_back();
        ans = min(ans, (long long)(s.size() - str.size()));
      }
    }
    cout << ans << endl;
  }
  return 0;
}
