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;
int64_t fnc(string s) {
  vector<int> v[10];
  for (int i = 0; i < s.size(); i++) v[s[i] - '0'].push_back(i);
  int64_t ans = 0;
  for (int i = 0; i < 10; i++) {
    ans = max(ans, int64_t(v[i].size()));
    for (int j = i + 1; j < 10; j++) {
      vector<bool> t;
      int64_t i1 = 0, j1 = 0;
      while (i1 < v[i].size() && j1 < v[j].size()) {
        if (v[i][i1] < v[j][j1]) {
          t.push_back(true);
          i1++;
        } else {
          t.push_back(false);
          j1++;
        }
      }
      while (i1 < v[i].size()) {
        t.push_back(true);
        i1++;
      }
      while (j1 < v[j].size()) {
        t.push_back(false);
        j1++;
      }
      if (t.size() == 0) continue;
      int64_t temp = 1;
      bool flag = !t[0];
      for (int k = 1; k < t.size(); k++) {
        if (t[k] == flag) {
          temp++;
          flag = !flag;
        }
      }
      ans = max(ans, (temp / 2) * 2);
    }
  }
  return s.size() - ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    cout << fnc(s) << endl;
  }
  return 0;
}
