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;
  for (int jw = 0; jw < t; jw++) {
    string s;
    cin >> s;
    map<int, int> m;
    int max = 0;
    int n = s.size();
    for (int i = 0; i < n; i++) {
      int we = s[i] - 48;
      m[we]++;
      if (m[we] > max) {
        max = m[we];
      }
    }
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        if ((i != j) && (m.find(i) != m.end()) && (m.find(j) != m.end())) {
          int weq = 0;
          for (int l = 0; l < n; l++) {
            int sd = s[l] - 48;
            if (weq % 2 == 0) {
              if (sd == i) {
                weq++;
              }
            } else {
              if (sd == j) {
                weq++;
              }
            }
          }
          if (weq % 2 != 0) {
            weq--;
          }
          if (weq > max) {
            max = weq;
          }
        }
      }
    }
    if (n > 2) {
      if (max < 2) {
        max = 2;
      }
    }
    cout << (n - max) << endl;
  }
}
