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;
int32_t main() {
  int tt;
  cin >> tt;
  while (tt--) {
    string s;
    cin >> s;
    int n = s.length();
    if (n == 2) {
      cout << 0 << "\n";
    } else {
      char ch = '0';
      int a[10];
      int mx = 0;
      for (int i = 0; i < 10; i++) {
        a[i] = count(s.begin(), s.end(), ch);
        ch++;
        mx = max(mx, a[i]);
      }
      for (int i = 0; i < 10; i++)
        for (int j = 0; j < 10; j++)
          if (i != j) {
            int cnt = 0;
            int x[2];
            x[0] = i, x[1] = j;
            int val = 0;
            for (int k = 0; k < n; k++) {
              int no = s[k] - '0';
              if (x[val] == no) {
                cnt++;
                val++;
                val %= 2;
              }
            }
            if (cnt & 1) cnt--;
            mx = max(mx, cnt);
          }
      cout << n - mx << "\n";
    }
  }
  return 0;
}
