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, i, j, k;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.length();
    map<int, int> m;
    for (i = 0; i <= n - 1; i++) {
      m[s[i] - 48]++;
    }
    int mini = n - 2, maxi = 0;
    for (i = 0; i <= 9; i++) {
      for (j = 0; j <= 9; j++) {
        if (i != j) {
          bool flag = true;
          int cnt = 0;
          for (k = 0; k <= n - 1; k++) {
            if (flag && s[k] == i + 48) {
              cnt++;
              flag = false;
            }
            if (!flag && s[k] == j + 48) {
              cnt++;
              flag = true;
            }
          }
          if (flag)
            mini = min(mini, n - cnt);
          else
            mini = min(mini, n - cnt + 1);
        }
      }
      maxi = max(maxi, m[i]);
    }
    mini = min(mini, n - maxi);
    m.clear();
    cout << mini << "\n";
  }
}
