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 T>
vector<T> readVector(int n) {
  vector<T> res(n);
  for (int i = 0; i < n; i++) cin >> res[i];
  return res;
}
void yes() {
  cout << "YES"
       << "\n";
}
void no() {
  cout << "NO"
       << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  cin >> tt;
  while (tt--) {
    string s;
    cin >> s;
    int n = s.length();
    int ans = n - 1;
    for (int a = 0; a <= 9; a++) {
      for (int b = 0; b <= 9; b++) {
        int cnt = 0;
        bool is = true;
        for (int i = 0; i < n; i++) {
          if (is) {
            if (s[i] - '0' == a)
              is = false;
            else
              cnt++;
          } else {
            if (s[i] - '0' == b)
              is = true;
            else
              cnt++;
          }
        }
        if ((n - cnt) % 2 == 0 || a == b) {
          ans = min(ans, cnt);
        }
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
