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
for nt in range(int(input())):
	s = list(input())
	n = len(s)
	c = Counter(s)
	m = 0
	for i in c:
		m = max(m,c[i])
	ans = min(n-2,n-m)
	count = {}
	numbers = "0123456789"
	for i in numbers:
		flag = False
		for j in range(n):
			if s[j]==i:
				found = {}
				flag = True
			elif flag:
				if s[j] not in found:
					found[s[j]] = 1
					if i+s[j] not in count:
						count[i+s[j]] = 1
					else:
						count[i+s[j]] += 1

	# print (n,count)
	m = 0
	for i in count:
		m = max(m,count[i])
	print (min(ans,n-2*m,n-2))