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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long t;
string s;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> t;
  while (t--) {
    cin >> s;
    long long n = s.size();
    long long maxi = n - 2;
    for (long long i = 0; i < 100; i++) {
      long long cur = 0, idx = 0;
      char c1, c2;
      if (i < 10) {
        c1 = '0';
        c2 = (char)(i + '0');
      } else {
        c2 = (char)(i % 10 + '0');
        c1 = (char)((long long)(i / 10) + '0');
      }
      while (idx < n) {
        if (cur % 2 == 0) {
          while (idx < n and s[idx] != c1) idx++;
          if (idx < n) cur++;
        } else {
          while (idx < n and s[idx] != c2) idx++;
          if (idx < n) cur++;
        }
        idx++;
      }
      if (cur % 2 == 1 and c1 != c2) continue;
      maxi = min(maxi, n - cur);
    }
    cout << maxi << "\n";
  }
  return 0;
}
