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.
# cook your dish here
from collections import defaultdict as dc
for _ in range(int(input())):
    s=input()
    #x={'0':0,'1':0,'2':0,'3':0,'4':0,'5':0,'6':0,'7':0,'8':0,'9':0}
    n=len(s)
    #for i in range(n):
        #x[s[i]]+=1
    p=0
    #for i in range(10):
        #p=max(p,x[chr(48+i)])
    for i in range(10):
        for j in range(10):
            f=0
            q=0
            if i!=j:
                for k in range(n):
                    if f==0:
                        if s[k]==chr(48+i):
                            f=1
                    elif f==1:
                        if s[k]==chr(48+j):
                            f=0
                            q+=1
                p=max(p,2*q)
            elif i==j:
                for k in range(n):
                    if s[k]==chr(48+i):
                        q+=1
                p=max(p,q)
                    
    if n==1:
        print(0)
    else:
        print(n-p)