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 sys import stdin
for _ in range(int(stdin.readline())):
    s=stdin.readline()
    n=len(s)
    if len(s)>2:
        m=2
    else:
        print(0)
        continue
    
    for i in range(10):
        m=max(m,s.count(str(i)))
    
    s1=s[0]
    for i in s[1:n]:
        if i!=s1[-1]:
            s1+=i
    s=s1
    
    for i in '0123456789':
        if i in s:
            j=s.index(i)
            d1={}
            d2={}
            for c in '0123456789':
                d1[c]=0
            temp=s[j+1:]
            for k in temp:
                if k!=i:
                    d2[k]=1
                else:
                    for c in d2.keys():
                        d1[c]+=1
                    d2={}
            m=max(m,max(d1.values())*2)
            
            j=n-s[::-1].index(i)
            d1={}
            d2={}
            for c in '0123456789':
                d1[c]=0
            temp=s[:j]
            for k in temp:
                if k!=i:
                    d2[k]=1
                else:
                    for c in d2.keys():
                        d1[c]+=1
                    d2={}
            m=max(m,max(d1.values())*2)
  
    print(n-m-1)