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 n = s.size();
  vector<long long> pos[10];
  for (long long i = 0; i < n; i++) {
    long long curr = s[i] - '0';
    pos[curr].emplace_back(i);
  }
  long long ans = 0;
  for (long long i = 0; i < 10; i++) {
    if (!pos[i].size()) {
      continue;
    }
    for (long long j = 0; j < 10; j++) {
      if (!pos[j].size()) {
        continue;
      }
      long long now = 0, lastTake = -1, made = 0;
      while (true) {
        if (!now) {
          auto it = upper_bound(pos[i].begin(), pos[i].end(), lastTake);
          if (it != pos[i].end()) {
            lastTake = *it;
          } else {
            break;
          }
        } else {
          auto it = upper_bound(pos[j].begin(), pos[j].end(), lastTake);
          if (it != pos[j].end()) {
            lastTake = *it;
          } else {
            break;
          }
        }
        now ^= 1;
        made++;
      }
      made = made - ((made & 1) && (i != j));
      ans = max(ans, made);
    }
  }
  cout << n - ans << '\n';
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}
