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;
void solve() {
  string s;
  cin >> s;
  long long int n = s.size();
  vector<long long int> v(100);
  long long int ans = n - 1;
  for (long long int i = 0; i < n; i++) {
    v[s[i] - '0']++;
  }
  for (long long int i = 0; i <= 9; i++) {
    ans = min(ans, n - v[i]);
  }
  for (long long int x = 0; x <= 9; x++) {
    for (long long int y = 0; y <= 9; y++) {
      if (x == y) continue;
      long long int d = 0, mark = 0;
      for (long long int i = 0; i < n; i++) {
        if (mark == 0 && s[i] - '0' == x) {
          d++;
          mark ^= 1;
        } else if (mark == 1 && s[i] - '0' == y) {
          d++;
          mark ^= 1;
        }
      }
      d = 2 * (d / 2);
      ans = min(ans, n - d);
    }
  }
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int q = 1;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}
