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 fun(string s, int a, int b) {
  int i, j, k, n = s.length();
  int cnt = 0;
  i = 0;
  j = 0;
  int cnta = 0, cntb = 0;
  for (i = 0; i < n; i++) {
    if (s[i] - '0' == a && cnta == 0) {
      cnta = 1;
    } else if (s[i] - '0' == b && cnta == 1) {
      cnta = 0;
      cnt++;
    }
  }
  return n - cnt * 2;
}
int fun2(string s) {
  int arr[10] = {0}, i;
  for (i = 0; i < 10; i++) arr[i] = 0;
  for (i = 0; i < s.length(); i++) {
    arr[s[i] - '0']++;
  }
  return *max_element(arr, arr + 10);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int i, j, k, n;
    string s;
    cin >> s;
    n = s.length();
    int mini = INT_MAX;
    i = 5, j = 1;
    for (i = 0; i < 10; i++) {
      for (j = 0; j < 10; j++) {
        string temp = s;
        int p = fun(temp, i, j);
        mini = min(mini, p);
      }
    }
    int a = fun2(s);
    cout << min(mini, n - a) << endl;
  }
}
