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;
int n, k, z;
void solve() {
  string s;
  cin >> s;
  n = s.size();
  map<char, int> cnt;
  for (auto c : s) cnt[c]++;
  int mx = -1, mx1 = -1;
  for (auto x : cnt) mx1 = max(mx1, x.second);
  if (mx1 == n) {
    cout << "0\n";
    return;
  }
  int mn = n - mx1;
  for (char c = '0'; c <= '9'; c++) {
    for (char z = '0'; z <= '9'; z++) {
      char curr = c, next = z;
      int count = 0;
      for (int i = 0; i < n; i++) {
        if (s[i] == curr) {
          count++;
          swap(curr, next);
        }
      }
      if (count & 1) count--;
      mn = min(mn, n - count);
    }
  }
  cout << mn << "\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;
}
