while(<>){
    chomp;
    next if $. == 1;
    if (/^(\d+)\.(\d+)\.(\d+)\.(\d+)$/){
    ($1 < 256 && $2 < 256 && $3 < 256 && $4 < 256) ? print "IPv4\n" : print "Neither\n";
        
    } elsif ( /^([a-f\d]{1,4}):([a-f\d]{1,4}):([a-f\d]{1,4}):([a-f\d]{1,4}):([a-f\d]{1,4}):([a-f\d]{1,4}):([a-f\d]{1,4}):([a-f\d]{1,4})$/ ){
        print "IPv6\n";
    } else {
        print "Neither\n";
    }

from ipaddress import ip_address

for _ in range(int(input())):
    try:
        addr = ip_address(input())
        print("IPv4" if addr.version == 4 else "IPv6")
    except ValueError:
        print("Neither")

import java.io.*;
import java.util.*;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class Solution {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int t = scanner.nextInt(); scanner.nextLine();
        String u = "(25[0-5]|2[0-4]\\d|1?\\d?\\d)"; //0 to 255
        String regex_ipv4 = String.format("^(%s\\.){3}%s$",u,u); //u.u.u.u
        String v = "[a-fA-F0-9]{1,4}"; //4 hexadecimal digits
        String regex_ipv6 = String.format("^(%s\\:){7}%s$",v,v); //v:v:v:v:v:v:v:v (8 groups)
        Pattern pattern_ipv4 = Pattern.compile(regex_ipv4);
        Pattern pattern_ipv6 = Pattern.compile(regex_ipv6);
        for(int i=0;i<t;i++)
        {   String str = scanner.nextLine();
            Matcher matcher_ipv4 = pattern_ipv4.matcher(str);
            Matcher matcher_ipv6 = pattern_ipv6.matcher(str);
            if (matcher_ipv4.find())    System.out.println("IPv4");
            else if (matcher_ipv6.find())   System.out.println("IPv6");
            else    System.out.println("Neither");
        }
    }
}

/*
---------- v = "[a-fA-F0-9]{1,4}" ---------------
4 hexadecimal digits.
Hexadecimal means either using a-f or 0-9
ff translates to 00ff. So no of digits can be either 1,2,3 or 4. So use {1,4}

---------- u = "(25[0-5]|2[0-4]\\d|1?\\d?\\d)" ---------------
Means 0 to 255
3 digit combinations: 
    25[0-5]  or 2[0-4][0-9] or 1[0-9][0-9] 
2 digit combinations:
    [0-9][0-9]
1 digit combinations:
    [0-9]
Combining 1 digit and 2 digit combinations: becomes: [0-9]?[0-9]
    Combine this with 1[0-9][0-9] : becomes : 1?[0-9]?[0-9]
So final set of combinations : becomes:
    25[0-5]  or 2[0-4][0-9] or 1?[0-9]?[0-9]


*/

import re
n = int(input())
for _ in range(n):
    address = input()
    regexv4 = r"^((25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])\.){3}(25[0-5]|2[0-4][0-9]|1[0-9]{2}|[1-9]?[0-9])$"
    regexv6 = r"^([a-f0-9]{1,4}:){7}([a-f0-9]{1,4})$"
    if re.match(regexv4, address):
        print("IPv4")
    elif re.match(regexv6, address):
        print("IPv6")
    else:
        print("Neither")

import re

pattern_ipv4 = r'^(\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])(\.(\d{1,2}|1\d{2}|2[0-4]\d|25[0-5])){3}$'
pattern_ipv6 = r'^([\da-f]{,4})(:([\da-f]{,4})){7}$'

for _ in range(int(input())):
    line = input()
    if re.match(pattern_ipv4, line):
        print('IPv4')
    elif re.match(pattern_ipv6, line):
        print('IPv6')
    else:
        print('Neither')