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,stdout
from collections import defaultdict as dd
from bisect import bisect_left as bl
input=stdin.readline
for _ in range(int(input())):
    a=list(input())
    a.pop()
    d=dd(list)
    for i in range(len(a)):
        d[int(a[i])].append(i)
    ans=0
    for i in d.keys():
        ans=max(ans,len(d[i]))
    x=list(d.keys())
    for i in range(len(x)):
        for j in range(len(x)):
            if i!=j:
                r=0
                s=bl(d[x[j]],d[x[i]][r])
                k=0
                while r<len(d[x[i]]) and s< len(d[x[j]]):
                    k+=2
                    r=bl(d[x[i]],d[x[j]][s])
                    if r<len(d[x[i]]):
                        s=bl(d[x[j]],d[x[i]][r])
                ans=max(ans,k)
                #print(x[i],x[j],k)
    stdout.write(str(len(a)-ans)+'\n')
                
            
            
            
            