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 sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
from collections import Counter as cc
from bisect import bisect as bi 

t,=I()
for _ in range(t):
	s=input().strip()
	p=cc(s)
	an=max(p.values())
	l=[[] for i in range(10)]
	for i in range(10):
		cr=chr(ord('0')+i)
		if s[0]==cr:
			l[i].append(0)
		for j in range(1,len(s)):
			if s[j]==cr and s[j-1]!=cr:
				l[i].append(j)
	for i in range(10):
		for j in range(i+1,10):
			x=list(l[i])
			y=list(l[j])
			if x and y:
				ct=0
				if x[0]>y[0]:
					x=l[j].copy()
					y=l[i].copy()
				if 1:
					kk=[]
					for pp in range(len(y)):
						st=bi(x,y[pp])
						if kk:
							if st!=kk[-1]:
								kk.append(st)
						else:
							kk.append(st)
					ct=2*len(kk)
				an=max(ct,an)
	print(len(s)-an)
