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() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long int t;
  cin >> t;
  long long int w = t;
  while (t--) {
    string s;
    cin >> s;
    long long int n = s.size();
    vector<long long int> occ(10, 0);
    vector<vector<long long int> > a(10, vector<long long int>(10, 0));
    for (long long int i = 0; i < (n); i++) {
      occ[s[i] - '0']++;
    }
    for (long long int i = 0; i < (10); i++) {
      set<long long int> se;
      long long int f = 0;
      for (long long int j = 0; j < (n); j++) {
        if (s[j] == ('0' + i)) {
          f = 1;
          vector<long long int> v((se).begin(), (se).end());
          long long int r = v.size();
          for (long long int k = 0; k < (r); k++) a[i][v[k]]++;
          se.clear();
        } else
          se.insert((long long int)(s[j] - '0'));
      }
    }
    long long int maxi = 0;
    for (long long int i = 0; i < (10); i++) {
      for (long long int j = 0; j < (10); j++) maxi = max(maxi, a[i][j]);
    }
    long long int po = *max_element((occ).begin(), (occ).end());
    cout << min(n - 2 * maxi, n - po) << "\n";
  }
}
