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 faster() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const long long INF = 1e18;
const long long MAXN = 1 << 20;
const long long MOD = 1e9 + 7;
const long long LOG = ceil(log(MAXN) / log(2));
const long double EPS = 1e-9;
const int dx[8] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dy[8] = {0, -1, 0, 1, 1, -1, 1, -1};
string s;
long long ans;
void solve() {
  cin >> s;
  ans = 0;
  for (char a = '0'; a <= '9'; ++a) {
    for (char b = '0'; b <= '9'; ++b) {
      long long res = 0;
      string q = "";
      q += a;
      q += b;
      int j = 0;
      for (int i = 0; i < s.size(); ++i) {
        if (s[i] == q[j]) {
          res++;
          j ^= 1;
        }
      }
      if (a != b && res & 1) res--;
      ans = max(res, ans);
    }
  }
  cout << s.size() - ans << '\n';
}
int main() {
  faster();
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  cerr << "\nTime elapsed " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  return 0;
}
