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.
import atexit, io, sys 

  
# A stream implementation using an in-memory bytes  
# buffer. It inherits BufferedIOBase. 

buffer = io.BytesIO() 

sys.stdout = buffer

  
# print via here 

@atexit.register 

def write(): 

    sys.__stdout__.write(buffer.getvalue()) 
for _ in range(input()):
    a=raw_input()
    b=[[] for j in range(10)]
    v=0
    for i in range(len(a)):
        b[int(a[i])].append(i)
        v=max(v,len(b[int(a[i])]))
    ans=len(a)-v
    
    for i in range(10):
        for j in range(10):
            if i!=j:
                p=b[i];q=b[j]
                m,n=0,0
                te=0
                pb=0
                while m<len(p) and n<len(q):
                    if p[m]<pb:
                        m+=1
                    elif p[m]<q[n]:
                        pb=q[n];m+=1;n+=1;te+=2
                
                    else:
                        n+=1;
                    
                
                ans=min(len(a)-te,ans)
    print ans
                
