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.
#Either of the form 'AAAA...' (all same chars)
#Or of the form 'ABAB...' (even number of 2 alternating characters)
from collections import Counter

t=int(input())
for _ in range(t):
    numList=[int(x) for x in list(input())]
    l=len(numList)
#    numIdx=[[] for _ in range(10)] #numIdx[0-9]=[indices of 0-9]
    
#    for i in range(l):
#        numIdx[numList[i]].append(i)
    
#    singleCharCase=l-max([len(x) for x in numIdx])
    singleCharCase=l-max(Counter(numList).values())
    
    #doubleCharCase
    doubleCharCase=l
    
    for i in range(9):
        for j in range(i+1,10): #i!=j
            A=i
            B=j
            cnts=0
            prev=-1
#            if len(numIdx[A])==0 or len(numIdx[B])==0: #0 AB can be formed
#                continue
            for x in numList:
                if x==A and prev!=A:
                    cnts+=1
                    prev=A
                elif x==B and prev!=B:
                    cnts+=1
                    prev=B
            cnts=(cnts//2)*2 #round down to nearest even integer
            doubleCharCase=min(doubleCharCase,l-cnts)
    
    print(min(singleCharCase,doubleCharCase))