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.
from sys import stdout, stdin
_input, _print = stdin.readline, stdout.write
_len, _range, _list, _int = len, range, list, int
from collections import Counter
from itertools import combinations


def solution():
    for _ in _range(_int(_input())):
        s = _list(_input())
        len_s = _len(s)
        z = Counter(s)
        ans = len_s - z.most_common(1)[0][1]

        pairs = combinations(_list(z), 2)
        for pair in pairs:
            n = 0
            c = None
            start_c = -1
            for i in s:
                if c is None:
                    if i == pair[0]:
                        c = 1
                        start_c = 0
                        continue
                    if i == pair[1]:
                        c = 0
                        start_c = 1
                        continue
                    n += 1
                else:
                    if i == pair[c]:
                        c += 1
                        c %= 2
                    else:
                        n += 1
            if start_c != c:
                n += 1
            if n < ans:
                ans = n
        print(ans-1)
solution()