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;
const int N = 1e6 + 5;
const int INF = 1e9 + 5;
const long long mod = 1e9 + 7;
const double pi = 3.1415926536;
int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, -1, 1};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int ans = 0;
    for (int i = 0; i < 10; i++) {
      for (int j = 0; j < 10; j++) {
        int cnt = 0;
        bool yes = true;
        for (int k = 0; k < s.size(); k++) {
          int cur = s[k] - '0';
          if (cur == i && yes)
            yes = false, cnt++;
          else if (cur == j && !yes)
            yes = true, cnt++;
        }
        if (i != j) cnt -= !yes;
        ans = max(ans, cnt);
      }
    }
    cout << s.size() - ans << endl;
  }
  return 0;
}
