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.
import sys
input = sys.stdin.readline
T = int(input())



for testcase in range(1,T+1):
    s = list(input())
    if s[-1] == '\n':
        s.pop()
    
    n = len(s)
    for i in range(n):
        s[i] = int(s[i])


    d = [0]*10
    for e in s:
        d[e] += 1

    res = n - max(d)
    for num in range(100):
        p,q = divmod(num,10)
        if p == q:
            continue
        flag = True
        tmp = 0
        for j in range(n):
            if flag and s[j] == p:
                flag = False
            elif (not flag) and s[j] == q:
                tmp += 2
                flag = True
        res = min(res,n-tmp)
    print(res)