#include <bits/stdc++.h>
using namespace std;
int appear[10];
int main(void) {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    memset(appear, 0, sizeof(appear));
    string s;
    cin >> s;
    if (s.size() == 2) {
      cout << 0 << '\n';
      continue;
    }
    int answer = s.size() - 2;
    int part = 0;
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        if (i == j) continue;
        int recur = 0;
        for (int k = 0; k < s.size(); k++) {
          if (recur == 0 || recur % 2 == 0) {
            if (s[k] - '0' == i) recur++;
          } else {
            if (s[k] - '0' == j) recur++;
          }
        }
        if (recur % 2 == 1) recur--;
        part = max(part, recur);
      }
    }
    if (part != 0) {
      int p = s.size() - part;
      answer = min(answer, p);
    }
    for (int i = 0; i < s.size(); i++) appear[s[i] - '0']++;
    int num = 0;
    for (int i = 0; i < 10; i++) num = max(num, appear[i]);
    num = s.size() - num;
    answer = min(num, answer);
    cout << answer << '\n';
  }
  return 0;
}

### Prompt

Please formulate 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 appear[10];
int main(void) {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    memset(appear, 0, sizeof(appear));
    string s;
    cin >> s;
    if (s.size() == 2) {
      cout << 0 << '\n';
      continue;
    }
    int answer = s.size() - 2;
    int part = 0;
    for (int i = 0; i <= 9; i++) {
      for (int j = 0; j <= 9; j++) {
        if (i == j) continue;
        int recur = 0;
        for (int k = 0; k < s.size(); k++) {
          if (recur == 0 || recur % 2 == 0) {
            if (s[k] - '0' == i) recur++;
          } else {
            if (s[k] - '0' == j) recur++;
          }
        }
        if (recur % 2 == 1) recur--;
        part = max(part, recur);
      }
    }
    if (part != 0) {
      int p = s.size() - part;
      answer = min(answer, p);
    }
    for (int i = 0; i < s.size(); i++) appear[s[i] - '0']++;
    int num = 0;
    for (int i = 0; i < 10; i++) num = max(num, appear[i]);
    num = s.size() - num;
    answer = min(num, answer);
    cout << answer << '\n';
  }
  return 0;
}

```