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 find_path(al, bl):
    len_path = 0
    az = bz = 0
    an, bn = len(al), len(bl)
    want_a = True
    ap = bp = -1
    while True:
        if want_a:
            for i in range(az, an):
                if al[i] > bp:
                    len_path += 1
                    ap = al[i]
                    az = i
                    want_a = False
                    break
            if want_a:
                return len_path
        else:
            for i in range(bz, bn):
                if bl[i] > ap:
                    len_path += 1
                    bp = bl[i]
                    bz = i
                    want_a = True
                    break
            if not want_a:
                return len_path

def main():
    t = int(input())
    for _ in range(t):
        s = input(); n = len(s)
        freq = {}; M = 0#; d = {}
        for i in range(n):
            if s[i] not in freq:
                freq[s[i]] = [i]
            else:
                freq[s[i]].append(i)
        for ca in "0123456789":
            if ca not in freq:
                continue
            for cb in "0123456789":
                if cb not in freq:
                    continue
                len_path = find_path(freq[ca], freq[cb])
                if len_path % 2 != 0:
                    len_path -= 1
                #d[(ca, cb)] = len_path
                M = max(M, len_path)
        max_freq = 0
        for val in freq.values():
            max_freq = max(max_freq, len(val))
        print(min(n - 2, n - max_freq, n - M))
        #print(d)
main()