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 int N = 1e6 + 5;
const int oo = 1e9 + 5;
const int INF = 1e9 + 5;
const long long int ooll = 1e18 + 5;
long long int v[N];
void solve() {
  string s;
  cin >> s;
  int n = s.size(), ans = n - 1;
  for (int odd = 0; odd < 10; ++odd) {
    for (int eve = 0; eve < 10; ++eve) {
      int need = 0, have = 0;
      for (auto it : s) {
        if (need) {
          if (eve == it - '0') {
            have++;
            need ^= 1;
          }
        } else {
          if (odd == it - '0') {
            have++;
            need ^= 1;
          }
        }
      }
      if (have & 1) {
        if (odd != eve) --have;
      }
      ans = min(ans, n - have);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}
