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.
#Consistency is the key.
#code by: amrit00
from sys import stdin,stdout
import math
##input=stdin.readline
##
##def print(x='',y='',end='\n'):
##    if y=='':
##        stdout.write(str(x)+end)
##    else:
##        stdout.write(str(x)+' '+str(y)+end)


def solve():
    ans=0
    s=input()
    l=len(s)
    
    d={}
    c=''
    for i in s:
        if i in d:
            d[i]+=1
            if d[i]>ans:
                ans=d[i]
        else:
            d[i]=1


    for i in range(10):
        for j in range(10):
            if i!=j:
                cur=0
                for k in range(l):
                    if cur&1:
                        if s[k]==str(i):
                            cur+=1
                    else:
                        if s[k]==str(j):
                            cur+=1
                cur-=cur%2
                ans=max(ans,cur)

    print(l-ans)
                            
                    

    
    
    
        
        
        
    
        
        


    
tt=1
tt=int(input())
for __ in range(tt):
    solve()
