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 findmax(vector<int> &a, vector<int> &b) {
  if (a.size() == 0 && b.size() == 0) return 0;
  if (a.size() == 0 || b.size() == 0) return 1;
  int i = 0, j = 0;
  int mxlena = 1;
  int al = a.size();
  int bl = b.size();
  int curr = a[0];
  int checkin = 1;
  while (i < al && j < bl) {
    if (checkin == 0) {
      int found = 0;
      while (i < al) {
        if (a[i] > curr) {
          curr = a[i];
          found = 1;
          break;
        }
        i++;
      }
      mxlena += found;
      if (found == 0) break;
      checkin = 1 - checkin;
    } else {
      int found = 0;
      while (j < bl) {
        if (b[j] > curr) {
          curr = b[j];
          found = 1;
          break;
        }
        j++;
      }
      mxlena += found;
      if (found == 0) break;
      checkin = 1 - checkin;
    }
  }
  return mxlena;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.length();
    vector<int> indices[10];
    for (int i = 0; i < n; i++) {
      indices[s[i] - '0'].push_back(i);
    }
    int mx = 0;
    for (int i = 0; i < 10; i++) {
      for (int j = 0; j < 10; j++) {
        int cur = findmax(indices[i], indices[j]);
        if (i != j) cur = cur & 1 ? cur - 1 : cur;
        mx = max(mx, cur);
      }
    }
    cout << n - mx << endl;
  }
  return 0;
}
