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 kehsihba() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long get(string s, string x) {
  long long ans = 0;
  long long j = 0;
  long long n = s.size(), m = x.size();
  for (long long i = 0; i < n; i++) {
    if (s[i] == x[j]) {
      j += 1;
    }
    if (j == m) {
      j = 0;
      ans += 1;
    }
  }
  return ans;
}
void solve() {
  string s;
  cin >> s;
  long long n = s.size();
  map<char, long long> m;
  map<string, long long> m1;
  string a = s[n - 1] + s.substr(0, n - 1);
  string b = s.substr(1, n - 1) + s[0];
  if (a == b) {
    cout << 0 << '\n';
    return;
  }
  for (auto i : s) {
    m[i] += 1;
  }
  long long ans = 0;
  for (auto i : m) {
    ans = max(i.second, ans);
  }
  if (ans == 1) {
    cout << n - 2 << '\n';
    return;
  }
  long long final = n - ans;
  for (long long i = 0; i < 10; i++) {
    for (long long j = 0; j < 10; j++) {
      string x = to_string(i) + to_string(j);
      long long c = get(s, x);
      final = min(final, n - 2 * c);
    }
  }
  cout << final << '\n';
}
int32_t main() {
  kehsihba();
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
