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;
char s[(200010)];
int main() {
  int t;
  cin >> t;
  while (t--) {
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    int ans = n - 2;
    for (int i = '0'; i <= '9'; i++) {
      for (int j = '0'; j <= '9'; j++) {
        int flag = 0, cnt = 0;
        for (int k = 1; k <= n; k++) {
          char goal = flag ? char(i) : char(j);
          if (goal == s[k]) {
            flag ^= 1;
            cnt++;
          }
        }
        if (i != j && cnt % 2) cnt--;
        ans = min(ans, n - cnt);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
