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 defaultdict, Counter
from math import log, ceil, floor

def solve():
	#n = int(input())
	#s = input()
	#n = len(s)
	#a, b= list(map(int, input().split()))
	s = list(input())
	# b = list(input())
	d = [0 for i in range(11)]
	#print(s)
	for i in s:
		d[int(i)] += 1
	ans1 = max(d)
	ans2 = 0
	for i in range(10) :
		for j in range(10):
			if i!=j:
				x = 1
				c = 0
				for k in s:
					if int(k) == i and x:
						c +=1
						x = 0
					elif int(k) == j and not x:
						c += 1
						x  =1
				ans2 = max(ans2, c)
	if ans2&1:
		ans2 -=1
	print(len(s) - max(ans2, ans1))

		


##################################################################
if __name__ == "__main__":
	t = 1
	t = int(input())
	for _ in range(t): solve()