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 math

def find_length(first, second):
	# finds longest string that can be formed from arr that has first in odd places and second in even places
	to_find = 0
	ops = [first, second]
	length = 0
	for char in arr:
		if (char == ops[to_find]):
			length += 1
			to_find = 1 - to_find
	if (to_find == 0):
		return (length)
	return (length-1)

num_cases = int(input())
for _ in range(num_cases):
	arr = list(map(int, input().strip()))
	n = len(arr)
	count = [0 for _ in range(10)]
	for i in arr:
		count[i] += 1
	max_odd = max(count)
	max_even = 0
	for first in range(0,10):
		for second in range(0,10):
			if (first == second):
				continue
			even = find_length(first, second)
			max_even = max(max_even, even)
	print(n - max(max_even, max_odd))
		
	
	



