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);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    map<char, int> mp;
    for (char c : s) {
      mp[c]++;
    }
    int mx = 0;
    for (auto i : mp) {
      mx = max(mx, i.second);
    }
    int mn = n - mx;
    for (int c = 0; c < 10; c++) {
      for (int cc = 0; cc < 10; cc++) {
        int cnt = 0;
        vector<int> cmp = {char(c + '0'), char(cc + '0')};
        int tog = 0;
        for (int i = 0; i < n; i++) {
          if (cmp[tog] == s[i])
            tog = tog ^ 1;
          else
            cnt++;
        }
        if ((n - cnt) % 2 == 1) cnt++;
        mn = min(mn, cnt);
      }
    }
    cout << mn << endl;
  }
  return 0;
}
