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;
void resolve() {
  string s;
  cin >> s;
  int mxL = 0;
  vector<int> q(10, 0);
  for (int i = 0; i < s.size(); i++) {
    q[s[i] - '0']++;
    if (q[s[i] - '0'] > mxL) mxL = q[s[i] - '0'];
  }
  for (char ft = '0'; ft <= '9'; ft++) {
    for (char sc = '0'; sc <= '9'; sc++) {
      if (sc == ft) continue;
      bool t = 0;
      int l = 0;
      for (int i = 0; i < s.size(); i++) {
        if (!t) {
          if (s[i] == ft) t = 1, l++;
        } else if (s[i] == sc)
          t = 0, l++;
      }
      mxL = max(mxL, l - l % 2);
    }
  }
  cout << s.size() - mxL;
}
int main() {
  srand(time(0) + clock());
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    resolve();
    cout << '\n';
  }
}
