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 t;
  cin >> t;
  while (t > 0) {
    --t;
    string s;
    cin >> s;
    int a[10] = {0};
    vector<int> b[10];
    for (int i = 0; i < s.size(); ++i) {
      ++a[s[i] - '0'];
      b[s[i] - '0'].push_back(i);
    }
    int ma = 0, ma1 = 0;
    for (int i = 0; i < 10; ++i) ma = max(ma, a[i]);
    for (int i = 0; i < 10; ++i)
      for (int j = 0; j < 10; ++j)
        if (i != j) {
          int k = 0, l = 0, r = 0;
          while (l < b[i].size() && r < b[j].size()) {
            if (k % 2 == 0) {
              if (k == 0)
                ++k;
              else {
                while (l < b[i].size() && b[i][l] < b[j][r]) ++l;
                if (l < b[i].size()) ++k;
              }
            } else {
              while (r < b[j].size() && b[i][l] > b[j][r]) ++r;
              if (r < b[j].size()) ++k;
            }
          }
          ma1 = max(ma1, k);
        }
    cout << s.size() - max(ma, ma1 - ma1 % 2) << endl;
  }
}
