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;
  scanf("%d", &t);
  while (t--) {
    char s[200001];
    scanf("%s", s);
    int n = strlen(s);
    int ans = n - 2;
    for (int i = '0'; i <= '9'; i++) {
      for (int j = '0'; j <= '9'; j++) {
        vector<char> arr;
        for (int k = 0; k < n; k++) {
          if (s[k] != i && s[k] != j) continue;
          if (i == j) {
            arr.push_back(s[k]);
          } else {
            if (arr.size() == 0 || s[k] != arr[arr.size() - 1])
              arr.push_back(s[k]);
          }
        }
        if (arr.size() < 2) continue;
        int abc = n - arr.size();
        if (i == j) {
          ans = min(ans, abc);
        } else {
          if (arr[0] == arr[arr.size() - 1])
            ans = min(ans, abc + 1);
          else
            ans = min(ans, abc);
        }
      }
    }
    printf("%d\n", ans);
  }
}
