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 solve() {
  string s;
  cin >> s;
  int n = s.size();
  int ans = 0;
  vector<int32_t> abc(10);
  int mxabc = 0;
  vector<vector<int32_t>> b(10);
  for (int i = 0; i < n; ++i) {
    abc[s[i] - '0']++;
    b[s[i] - '0'].push_back(i);
    mxabc = max(mxabc, abc[s[i] - '0']);
  }
  for (int i = 0; i < 10; ++i) {
    for (int j = 0; j < 10; ++j) {
      if (i == j) continue;
      int cnt = 0;
      for (int ii = 0, jj = 0; ii < b[i].size() && jj < b[j].size();) {
        if (b[i][ii] < b[j][jj]) {
          while (ii < b[i].size() && b[i][ii] < b[j][jj]) ii++;
          cnt += 2;
        } else {
          jj++;
        }
      }
      ans = max(ans, cnt);
    }
  }
  cout << min(n - ans, n - mxabc) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t_;
  cin >> t_;
  while (t_--) solve();
  return 0;
}
