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);
  int T;
  cin >> T;
  for (int t = 0; t < T; ++t) {
    string s;
    cin >> s;
    int n = s.size();
    vector<int> v;
    for (int i = 0; i < n; i++) {
      v.push_back(s[i] - '0');
    }
    int minlen = INT_MAX;
    for (int t1 = 0; t1 <= 9; t1++) {
      for (int t2 = 0; t2 <= 9; t2++) {
        int count = 0, flag = 0;
        for (int i = 0; i < n; i++) {
          if (flag == 0 && v[i] == t1) {
            flag = 1;
            count++;
          } else {
            if (flag == 1 && v[i] == t2) {
              flag = 0;
              count++;
            }
          }
        }
        int ans = 0;
        if (t1 == t2) {
          ans = n - count;
        } else {
          ans = n - count + flag;
        }
        minlen = min(ans, minlen);
      }
    }
    cout << minlen << "\n";
  }
}
