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

t = int(input())

for _ in range(t):

    ar = list(map(int, list(input())))
    n = len(ar)
    cntr = Counter(ar)
    Set = list(cntr.keys())

    if len(cntr)==1:
        print(0)

    elif len(cntr)==n:
        print(n-2)

    else:
        
        checkAr = [[-1 for i in range(10)] for j in range(10)]
        countAr = [[0 for i in range(10)] for j in range(10)]
        mx = max(cntr.values())

        for idx in range(n):

            if idx>0 and ar[idx]==ar[idx-1]:
                continue
            
            char = ar[idx]
            for i in Set:
                if char>i:
                    if checkAr[char][i] != char:
                        checkAr[char][i] = char
                        countAr[char][i] += 1
                        mx = max(countAr[char][i],mx)
                elif checkAr[i][char] != char:
                    checkAr[i][char] = char
                    countAr[i][char] += 1
                    mx = max(countAr[i][char],mx)

        if mx!=max(cntr.values()) and mx%2==1:
            mx = mx-1
        output = n-mx
        print(output)