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 solve() {
  string S;
  cin >> S;
  int n = (int)S.size();
  int score = 0;
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      if (i == j) continue;
      int prev = -1;
      int first = -1;
      int tmp_score = 0;
      for (char c : S) {
        if (c - '0' == i) {
          if (prev == i)
            continue;
          else {
            tmp_score++;
            prev = i;
            if (first == -1) first = i;
          }
        } else if (c - '0' == j) {
          if (prev == j)
            continue;
          else {
            tmp_score++;
            prev = j;
            if (first == -1) first = j;
          }
        }
      }
      if (first == prev) tmp_score--;
      score = max(tmp_score, score);
    }
  }
  map<char, int> C;
  int mx = 0;
  for (char c : S) {
    C[c]++;
    mx = max(mx, C[c]);
  }
  cout << n - max(score, mx) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) solve();
}
