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 sys import stdin
input = stdin.readline


def answer():

    count = [0] * (10)

    for i in s:
        count[int(i)] += 1

    ans = n - max(count)

    for v1 in range(10):

        for v2 in range(10):

            x , take = 0 , 0
            for i in range(n):

                if(take == 0):
                    if(int(s[i]) == v1):
                        x += 1
                        take = 1
                else:
                    if(int(s[i]) == v2):
                        x += 1
                        take = 0

            x -= take
            ans = min(ans , n - x)

    return ans

for T in range(int(input())):

    s = input().strip()
    n = len(s)
    
    
    print(answer())
