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 main() {
  int t;
  cin >> t;
  while (t--) {
    int n, tmp = 0, ans = 0;
    string s;
    cin >> s;
    n = s.length();
    for (char i = '0'; i <= '9'; ++i) {
      for (char j = '0'; j <= '9'; ++j) {
        for (int x = 0; x < n; ++x) {
          if (tmp & 1) {
            if (s[x] == i) ++tmp;
          } else if (!(tmp & 1)) {
            if (s[x] == j) ++tmp;
          }
        }
        if (i != j) {
          if (tmp & 1) --tmp;
        }
        ans = max(ans, tmp);
        tmp = 0;
      }
    }
    cout << n - ans << "\n";
  }
}
