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 long long N = 1000000007;
void solve() {
  string s;
  cin >> s;
  long long ans = INT_MAX;
  for (long long i = 0; i <= 9; i++) {
    long long count = 0;
    for (long long j = 0; j < (long long)s.size(); j++) {
      if (s[j] == (char)('0' + i)) count++;
    }
    ans = min(ans, (long long)s.size() - count);
  }
  if (ans == 0) {
    cout << 0 << endl;
    return;
  }
  for (long long i = 0; i <= 9; i++) {
    for (long long j = 0; j <= 9; j++) {
      string temp;
      if (i == j) continue;
      temp += (char)('0' + i);
      temp += (char)('0' + j);
      long long count = 0;
      long long n = (long long)s.size();
      long long curr = 0;
      for (long long k = 0; k < n; k++) {
        if (s[k] == temp[curr]) {
          if (curr == 1) {
            count++;
          }
          curr = (curr + 1) % 2;
        }
      }
      ans = min(ans, (long long)s.size() - 2 * count);
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
