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.
def count_alternation(s, a, b):
    n = len(s)

    # Make "a" appears before "b" (WLOG)
    for i in range(n):
        if s[i] == a:
            break
        if s[i] == b:
            a, b = b, a
            break

    # Find right most s[i1] == "b"
    i1 = 0
    for i in range(n)[::-1]:
        if s[i] == b:
            i1 = i
            break

    # Count alternation
    flip = {a: b, b: a}
    which = a
    count = 0
    for i in range(i1 + 1):
        if s[i] == which:
            if which == b:
                count += 1
            which = flip[which]

    return 2 * count


def solve(s):
    # Convert to numeric
    s = [ord(c) - ord("0") for c in s]

    k = 10  # 10 types of element
    n = len(s)
    counts = [0] * k
    for i in range(n):
        counts[s[i]] += 1

    # 1. Single type solution
    result = max(counts)

    # 2. Two types solution by brute force
    for a in range(k):
        for b in range(a + 1, k):
            if counts[a] == 0 or counts[b] == 0:
                continue
            result = max(result, count_alternation(s, a, b))

    return n - result


def main(istr, ostr):
    t = int(istr.readline())
    for _ in range(t):
        s = istr.readline().strip()
        result = solve(s)
        print(result, file=ostr)


if __name__ == "__main__":
    import sys

    main(sys.stdin, sys.stdout)
