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>
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
using namespace std;
int main() {
  int nq;
  cin >> nq;
  while (nq--) {
    string str;
    cin >> str;
    auto arr = vector<int>(str.size());
    for (int i = 0; i < arr.size(); i++) arr[i] = str[i] - '0';
    int ans = arr.size() - 2;
    for (int i = 0; i < 10; i++) {
      for (int j = 0; j < 10; j++) {
        int found = 0;
        for (int k = 0; k < arr.size(); k++) {
          if (arr[k] == (found % 2 ? i : j)) found++;
        }
        if (found % 2 && i != j) found--;
        int cur = arr.size() - found;
        ans = min(ans, cur);
      }
    }
    std::cout << ans << "\n";
  }
  return 0;
}
