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 functools import reduce
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase

input = lambda: sys.stdin.readline().rstrip("\r\n")
def value():return tuple(map(int,input().split()))
def arr():return [int(i) for i in input().split()]
def sarr():return [int(i) for i in input()]
def inn():return int(input())
#----------------------------CODE------------------------------#
t=int(input())
while t:
    t-=1
    a=sarr()
    n=len(a)
    ln = len(set(a))
    if(ln==1):
        print(0)
    elif(ln==n):
        print(n-2)
    else:
        d=defaultdict(int)
        for i in a:
            if(i not in d):
                d[i]=1
            else:
                d[i]+=1
        mx = max(d.values())
        ans=1000000
        for i in range(10):
            for j in range(10):
                if(i!=j):
                    c=0
                    id=0
                    while(id<n):
                        while(id<n and a[id]!=i):
                            id+=1
                        while(id<n and a[id]!=j):
                            id+=1
                        if(id==n):
                            break
                        c+=2
                        ans=min(ans,n-c)
        print(min(ans,n-mx))
