#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    vector<int> has(10, 0);
    for (int i = 0; i < n; i++) {
      has[s[i] - '0']++;
    }
    int count = 0;
    for (int i = 0; i < 10; i++) {
      if (has[i] > 0) count++;
    }
    if (count == 1)
      cout << "0\n";
    else {
      int maxi = *max_element(has.begin(), has.end());
      int ans = s.size() - maxi;
      for (int i = 0; i <= 9; i++) {
        for (int j = 0; j <= 9; j++) {
          int f = 0;
          int res = 0;
          for (int k = 0; k < n; k++) {
            if (f == 0 && s[k] == i + '0') {
              res++;
              f ^= 1;
            } else if (f == 1 && s[k] == j + '0') {
              res++;
              f ^= 1;
            } else
              continue;
          }
          if (res % 2) res--;
          ans = min(ans, n - res);
        }
      }
      cout << ans << endl;
    }
  }
}

### Prompt

Create a solution in CPP for the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int n = s.size();
    vector<int> has(10, 0);
    for (int i = 0; i < n; i++) {
      has[s[i] - '0']++;
    }
    int count = 0;
    for (int i = 0; i < 10; i++) {
      if (has[i] > 0) count++;
    }
    if (count == 1)
      cout << "0\n";
    else {
      int maxi = *max_element(has.begin(), has.end());
      int ans = s.size() - maxi;
      for (int i = 0; i <= 9; i++) {
        for (int j = 0; j <= 9; j++) {
          int f = 0;
          int res = 0;
          for (int k = 0; k < n; k++) {
            if (f == 0 && s[k] == i + '0') {
              res++;
              f ^= 1;
            } else if (f == 1 && s[k] == j + '0') {
              res++;
              f ^= 1;
            } else
              continue;
          }
          if (res % 2) res--;
          ans = min(ans, n - res);
        }
      }
      cout << ans << endl;
    }
  }
}

```