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;
const long long M = 1000000007;
long long compare(pair<long long, long long> &a,
                  pair<long long, long long> &b) {
  return a.first > b.first;
}
long long compare2(pair<long long, long long> &a,
                   pair<long long, long long> &b) {
  return a.second < b.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T;
  cin >> T;
  while (T--) {
    long long ans = 0;
    string s;
    cin >> s;
    for (long long i = 0; i < 10; i++) {
      for (long long j = 0; j < 10; j++) {
        char ch1 = '0' + i;
        char ch2 = '0' + j;
        char cur = ch1;
        long long tmp = 0;
        for (long long k = 0; k < s.length(); k++) {
          if (s[k] == cur) {
            tmp++;
            if (cur == ch1)
              cur = ch2;
            else
              cur = ch1;
          }
        }
        if (ch1 == ch2) {
          ans = max(ans, tmp);
        } else {
          if (tmp % 2 != 0) tmp--;
          ans = max(ans, tmp);
        }
      }
    }
    ans = (s.length()) - ans;
    cout << ans << endl;
  }
}
