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;
long long i, j, k;
struct cmp {
  bool operator()(long long const& p1, long long const& p2) { return p1 > p2; }
};
void Anon_mouS() {
  string s;
  cin >> s;
  map<long long, long long> m;
  long long n = s.length();
  long long ans = 0;
  for (i = 0; i < n; i++) {
    m[s[i] - '0']++;
    ans = max(ans, m[s[i] - '0']);
  }
  for (i = 0; i <= 9; i++) {
    for (j = 0; j <= 9; j++) {
      if (i == j) continue;
      long long c = 0;
      for (k = 0; k < n; k++) {
        if (c % 2 == 1 and s[k] - '0' == j) c++;
        if (c % 2 == 0 and s[k] - '0' == i) c++;
      }
      c -= c % 2;
      ans = max(ans, c);
    }
  }
  cout << n - ans << '\n';
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long q = 1;
  cin >> q;
  for (long long it = 1; it <= q; it++) {
    Anon_mouS();
  }
  return 0;
}
