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() {
  int test;
  cin >> test;
  while (test--) {
    string s;
    cin >> s;
    int ans = INT_MAX;
    unordered_map<char, int> mp;
    int n = s.size();
    for (int i = 0; i < s.size(); i++) {
      mp[s[i]]++;
    }
    for (auto x : mp) {
      ans = min(ans, n - x.second);
    }
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        int val = i;
        int ans1 = 0;
        for (int k = 0; k < s.size(); k++) {
          int value = s[k] - '0';
          if (value == val) {
            if (val == i) {
              val = j;
              ans1++;
            } else if (val == j && j != i) {
              val = i;
              ans1++;
            }
          }
        }
        ans1 = (ans1 / 2) * 2;
        ans = min(ans, n - ans1);
      }
    }
    cout << ans << endl;
  }
  return 0;
}
