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.
#input template
from sys import stdin, stdout
import math
cin = stdin.readline
cout = stdout.write
mp = lambda: list(map(int, cin().split()))

def chars(): #function for taking string input as character array since string in python is immutable
    s = cin()
    return(list(s[:len(s) - 1]))

#print list	
def pl(a):	
	for val in a:
		cout(str(val) + ' ')
	cout('\n')
		
#main
from collections import Counter
t, = mp()
for _ in range(t):	
	d = cin()[:-1]#chars()
	#s = Counter(d)
	ln = len(d)
	
	s = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
	
	ans = 0
	for ch in s:
		x = d.find(ch)
		if x < 0: continue
		for ch2 in s:
			f = temp = 0
			if ch == ch2:
				temp = 1
			for i in range(x+1, ln):
				if ch == ch2:
					if d[i] == ch:
						temp += 1
				elif d[i] == ch2 and not f:
					f = 1
					temp += 1
				elif d[i] == ch and f:
					f = 0
			if ch != ch2:
				temp *= 2
			#print(ch, ch2, temp)
			ans = max(temp, ans)
	
	cout(str(ln - max(ans, 2)) + '\n')