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 lINF = 1e18 + 7;
const int INF = 1e9 + 7;
template <class T>
ostream& operator<<(ostream& s, const vector<T>& v) {
  for (size_t i = 0; i < v.size(); i++) s << v[i] << ' ';
  return s;
}
template <class T>
istream& operator>>(istream& s, vector<T>& v) {
  for (size_t i = 0; i < v.size(); i++) s >> v[i];
  return s;
}
void solve(int CASE) {
  string s;
  cin >> s;
  int ans = (int)(s).size();
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      bool cur = true;
      int cnt = 0;
      for (char c : s) {
        if (cur && c - '0' == i || !cur && c - '0' == j) {
          cnt++;
          cur = !cur;
        }
      }
      if (cnt % 2 == 0 || i == j) ans = min(ans, (int)(s).size() - cnt);
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TEST = 1;
  cin >> TEST;
  for (int i = 1; i <= TEST; i++) solve(i);
}
