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 MM = 2e5 + 5;
string s;
int a[MM];
void solve(int tc) {
  cin >> s;
  int n = s.size();
  for (int i = 0; i < n; i++) a[i + 1] = s[i] - '0';
  int ans = 0;
  for (int x = 0; x < 10; x++) {
    for (int y = 0; y < 10; y++) {
      int cnt = 0, len = 0;
      for (int i = 1; i <= n; i++) {
        if (a[i] == x && cnt % 2 == 0) {
          cnt++;
          len++;
        } else if (a[i] == y && cnt % 2 == 1) {
          cnt++;
          len++;
        }
      }
      if (cnt % 2 == 1 && x != y) len--;
      ans = max(ans, len);
    }
  }
  cout << min(n - ans, n - 2) << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tc = 1;
  cin >> tc;
  for (int t = 1; t <= tc; t++) solve(t);
  return 0;
}
