#include <bits/stdc++.h>
using namespace std;
char s[200001];
bool sign;
int main() {
  int t, n, ans, now;
  scanf("%d", &t);
  while (t--) {
    ans = INT_MAX, scanf("%s", s), n = strlen(s);
    for (char a = '0'; a <= '9'; a++) {
      now = 0;
      for (int i = 0; i < n; i++)
        if (s[i] != a) now++;
      ans = min(ans, now);
    }
    for (char a = '0'; a <= '9'; a++)
      for (char b = '0'; b <= '9'; b++)
        if (a != b) {
          now = 0, sign = true;
          for (int i = 0; i < n; i++)
            if ((sign && s[i] == a) || (!sign && s[i] == b))
              sign = !sign;
            else
              now++;
          if ((n - now) & 1) now++;
          ans = min(ans, now);
        }
    printf("%d\n", ans);
  }
  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;
char s[200001];
bool sign;
int main() {
  int t, n, ans, now;
  scanf("%d", &t);
  while (t--) {
    ans = INT_MAX, scanf("%s", s), n = strlen(s);
    for (char a = '0'; a <= '9'; a++) {
      now = 0;
      for (int i = 0; i < n; i++)
        if (s[i] != a) now++;
      ans = min(ans, now);
    }
    for (char a = '0'; a <= '9'; a++)
      for (char b = '0'; b <= '9'; b++)
        if (a != b) {
          now = 0, sign = true;
          for (int i = 0; i < n; i++)
            if ((sign && s[i] == a) || (!sign && s[i] == b))
              sign = !sign;
            else
              now++;
          if ((n - now) & 1) now++;
          ans = min(ans, now);
        }
    printf("%d\n", ans);
  }
  return 0;
}

```