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 functools import lru_cache
from sys import stdin, stdout
import sys
from math import *

# from collections import deque
# sys.setrecursionlimit(int(1e3))
# input = stdin.buffer.readline
# print = stdout.write
# dp=[-1]*100000

for __ in range(int(input())):
	s=input()
	rslt=inf
	for i in range(10):
		for j in range(10):
			x=str(i)
			y=str(j)
			ans=0
			n=x
			for k in range(len(s)):
				if(s[k]==n):
					ans+=1
					if(n==x):
						n=y
					else:
						n=x
			if(ans%2==1 and x!=y):
				ans-=1
			rslt=min(len(s)-ans,rslt)
	print(rslt)



