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 Counter
from operator import itemgetter 

for x in range(int(input())):
    s =input()
    n=len(s)
    p={}
    for i in range(n):
        if s[i] not in p:
            p[s[i]]=i
    d=Counter(s)
    j=sorted(d.items(),key=itemgetter(1), reverse=True)
    #print(p,j)
    if j[0][1]>n//2:
        print(n-j[0][1])
    else:
        mx=j[0][1]
        for i in range(len(j)-1):
            
            
            for u in range(i+1,len(j)):
                if p[j[i][0]]<p[j[u][0]]:
                    t=0
                else:
                    t=1
                r=t
                l=0
                for k in range(n):
                    #print(l)
                    if s[k]==j[i][0] and t==0:
                        l+=1
                        t=1-t
                    elif s[k]==j[u][0] and t==1:
                        l+=1
                        t=1-t
                if t!=r:
                    l-=1
                
                mx=max(mx,l)
                        
                        
                
                
            
        print(n-mx)