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 scan_arr1(long long int arr[], long long int n, long long int start = 0) {
  for (long long int i = start; i < n + start; i++) cin >> arr[i];
}
void scan_arr2(long long int arr[1010][1010], long long int n, long long int m,
               long long int start = 0) {
  for (long long int i = start; i < n + start; i++)
    for (long long int j = start; j < m + start; j++) cin >> arr[i][j];
}
void print_arr1(long long int arr[], long long int n, long long int start = 0) {
  for (long long int i = start; i < n + start; i++) cout << arr[i] << " ";
  cout << "\n";
}
void print_arr2(long long int arr[1010][1010], long long int n, long long int m,
                long long int start = 0) {
  for (long long int i = start; i < n + start; i++) {
    for (long long int j = start; j < m + start; j++) cout << arr[i][j] << " ";
    cout << "\n";
  }
}
int main() {
  long long int t, n, i, j, k;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    n = s.length();
    long long int res = -1, cnt = 0;
    for (i = 0; i < 10; i++)
      for (j = 0; j < 10; j++) {
        bool f = 0;
        cnt = 0;
        for (k = 0; k < n; k++) {
          if (f == 0 && (s[k] - '0') == i) {
            cnt++;
            f = 1;
          } else if (f == 1 && (s[k] - '0') == j) {
            cnt++;
            f = 0;
          }
        }
        if (cnt % 2 && i != j) cnt--;
        res = max(res, cnt);
      }
    n -= res;
    cout << n << "\n";
  }
}
