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;
long long const mo = 1e9 + 7;
long double const pi = 2.00 * asin(1.00);
long long mod(long long k) { return (k + mo) % mo; }
long long addmod(long long a, long long b) { return mod(mod(a) + mod(b)); }
long long mulmod(long long a, long long b) { return mod(mod(a) * mod(b)); }
long long submod(long long a, long long b) {
  return mod(a % mo - b % mo + mo) % mo;
}
void solve() {
  string s;
  cin >> s;
  int res = 0;
  vector<int> v(10, 0);
  for (int i = 0; i < s.size(); ++i) {
    v[s[i] - '0']++;
  }
  int res1 = 1;
  for (int i = 0; i <= 9; ++i) res1 = max(res1, v[i]);
  res1 = s.size() - res1;
  for (char i = '0'; i <= '9'; ++i) {
    for (char j = i + 1; j <= '9'; ++j) {
      int noi = 0, noj = 0;
      bool x = true, y = true;
      for (int k = 0; k < s.size(); ++k) {
        if (x && s[k] == i) {
          ++noi;
          x = false;
          y = true;
        }
        if (y && s[k] == j) {
          ++noj;
          y = false;
          x = true;
        }
      }
      res = max(res, min(noi, noj));
    }
  }
  cout << min(res1, (int)s.size() - 2 * res) << "\n";
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout << fixed;
  cout.precision(15);
  ;
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; ++i) {
    solve();
  }
  return 0;
}
