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 stdin


def inp():
    return stdin.buffer.readline().rstrip().decode("utf8")


def itg():
    return int(stdin.buffer.readline())


def mpint():
    return map(int, stdin.buffer.readline().split())


# ############################## import
# ############################## main
from collections import Counter
for case in range(1, itg() + 1):
    s = inp()
    n = len(s)
    ans = n - Counter(s).most_common(1)[0][1]
    for a in range(10):
        a = str(a)
        for b in range(10):
            b = str(b)
            if a == b:
                continue
            count = 0
            next_is_a = True
            for c in s:
                if next_is_a:
                    if c == a:
                        next_is_a = not next_is_a
                        continue
                elif c == b:
                    next_is_a = not next_is_a
                    continue
                count += 1
            if (n-count) & 1:  # is odd, ababa
                count += 1
            ans = min(ans, count)
    print(ans)
"""
odd
01234
12340
40123
1=4=3=0=2

even
012345
123450
501234
1=5=3
2=0=4
ababab
"""

# Please check!
