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 collections import Counter
from itertools import combinations
for _ in range(int(input())):
    a = list(input())
    n = len(a)
    dic = Counter(a)
    b = list(sorted(dic.items(), key=lambda x: x[1], reverse=True))
    ans = b[0][1]
    if len(b) == 1: print(0)
    else:

        for xxx in combinations(dic.keys(), 2):
            i, j = xxx[0], xxx[1]
            if dic[i]+dic[j] < ans:
                continue

            c = ''.join(list(filter(lambda x: x in {i, j}, a)))
            ans=max(ans, 2*c.count(i+j), 2*c.count(j+i))
        print(n-2) if ans==1 else print(n-ans)
