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 check(vector<int>& a, vector<int>& b) {
  int count1(0), count2(0);
  bool hasa(false);
  int i(0), j(0);
  int last(-1);
  while (i < a.size() && j < b.size()) {
    if (hasa) {
      if (b[j] > last) {
        count1 += 2;
        i++;
        last = b[j];
        hasa = false;
      }
      j++;
    } else {
      if (a[i] > last) {
        hasa = true;
        last = a[i];
      } else
        i++;
    }
  }
  bool hasb(false);
  i = 0;
  j = 0;
  last = -1;
  while (i < a.size() && j < b.size()) {
    if (hasb) {
      if (a[i] > last) {
        count2 += 2;
        j++;
        last = a[i];
        hasb = false;
      }
      i++;
    } else {
      if (b[j] > last) {
        hasb = true;
        last = b[j];
      } else
        j++;
    }
  }
  return max(count1, count2);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    vector<int> v[10];
    for (int i = 0; i < s.size(); i++) v[s[i] - '0'].emplace_back(i);
    int good(0);
    for (int i = 0; i < 10; i++) {
      good = max(good, int(v[i].size()));
      for (int j = i + 1; j < 10; j++)
        if (v[i].size() && v[j].size()) {
          good = max(good, check(v[i], v[j]));
        }
    }
    cout << s.size() - good << endl;
  }
  return 0;
}
