#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    string s;
    cin >> s;
    int ans = 1000000000;
    for (int i = 0; i < 10; i++) {
      for (int j = 0; j < 10; j++) {
        int comp = 0;
        int next = i;
        for (int k = 0; k < s.length(); k += 1) {
          if (s[k] - '0' != next)
            comp += 1;
          else {
            if (next == i) {
              next = j;
            } else {
              next = i;
            }
          }
        }
        if (i != j && (s.length() - comp) % 2 == 1) {
          comp++;
        }
        ans = min(ans, comp);
      }
    }
    cout << ans << endl;
  }
}

### Prompt

Generate a CPP solution to 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 ans = 1000000000;
    for (int i = 0; i < 10; i++) {
      for (int j = 0; j < 10; j++) {
        int comp = 0;
        int next = i;
        for (int k = 0; k < s.length(); k += 1) {
          if (s[k] - '0' != next)
            comp += 1;
          else {
            if (next == i) {
              next = j;
            } else {
              next = i;
            }
          }
        }
        if (i != j && (s.length() - comp) % 2 == 1) {
          comp++;
        }
        ans = min(ans, comp);
      }
    }
    cout << ans << endl;
  }
}

```