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 *
import sys  

# "". join(strings)  
    
def ri():
    return int(input())
 
def rl():
    return list(map(int, input().split()))
    

t=ri()
for _ in range(t):

	n=input()
	l = set([x for x in n])
	cc = Counter(n)
	
	ans = len(n)-max(cc.values())
	
	val=0
	for i in l:
		for j in l:
			if j!=i:
				
				cnt=0
				k=0
				while k<len(n):
					if n[k]==i:
						z=k+1
						while z<len(n):
							if n[z]==j:
								cnt+=1
								k=z
								break
							z+=1
					k+=1
				val=max(cnt,val)
	ans = min(ans, len(n)-2*val)
	print(ans)

