Code Forces 21 A(模拟)

Jabber ID on the national Berland service «Babber» has a form <username>@<hostname>[/resource], where

<username> — is a sequence of Latin letters (lowercase or uppercase), digits or underscores characters «_», the length of<username> is between 1 and 16, inclusive.
<hostname> — is a sequence of word separated by periods (characters «.»), where each word should contain only characters allowed for <username>, the length of each word is between 1 and 16, inclusive. The length of <hostname> is between 1 and 32, inclusive.
<resource> — is a sequence of Latin letters (lowercase or uppercase), digits or underscores characters «_», the length of<resource> is between 1 and 16, inclusive.

The content of square brackets is optional — it can be present or can be absent.

There are the samples of correct Jabber IDs: mike@codeforces.com, 007@en.codeforces.com/contest.

Your task is to write program which checks if given string is a correct Jabber ID.

Input

The input contains of a single line. The line has the length between 1 and 100 characters, inclusive. Each characters has ASCII-code between 33 and 127, inclusive.

Output

Print YES or NO.

Examples

input
mike@codeforces.com

output
YES

input
john.smith@codeforces.ru/contest.icpc/12

output
NO
直接模拟#include<stdio.h>
#include<string.h>
bool panduan(char c)
{
	if((c!='_')&&(c<'a'||c>'z')&&(c<'A'||c>'Z')&&(c<'0'||c>'9'))return false;
	else return true;
}
int main()
{
	char s[100+10];
	gets(s);



	int len=strlen(s);
	char user[100];
	char host[100];
	char res[100];
	int i,len1,len2,len3;
	int ans=1;
	//user
	for(i=0;i<len;i++)
	{
		if(s[i]=='@')
		{
			len1=i;
			break;
		}
		user[i]=s[i];
		if(i==len-1&&s[i]!='@')ans=0;
	}
	if(s[len-1]=='@')
        ans=0;
	if(ans==1)
	{
		if(len1<1||len1>16)ans=0;
		else
		{
			for(i=0;i<len1;i++)
			{
				if(!panduan(user[i]))
				{
					ans=0;
					break;
				}
			}
		}
	}
	//host
	for(i=len1+1;i<len;i++)
	{
		if(s[i]=='/')
		{
			len2=i-len1-1;
			break;
		}
		host[i-len1-1]=s[i];
		if(i==len-1&&s[i]!='/')
		{
			len2=len-len1-1;
		}
	}
	if(ans==1)
	{
		if(len1<1||len1>32)ans=0;
		else
		{
			int sum=0;
			for(i=0;i<len2;i++)
			{
				if(!panduan(host[i]))
				{
					if(host[i]=='.')
                    {
                        if(sum>16||sum<1||(i==len2-1))
				        {
					       ans=0;break;
				        }
                        sum=0;
                    }
					else ans=0;
				}
				else sum++;
				if(ans==0)
                    break;

			}
		}
	}
	//res
	if(s[len-1]=='/')
        ans=0;
	if(len1+len2+2<len&&ans==1)
	{
		for(i=len1+len2+2;i<len;i++)
		res[i-len1-len2-2]=s[i];
		len3=len-2-len1-len2;
		if(len3<1||len3>16)ans=0;
		else{
		for(i=0;i<len3;i++)
		{
			if(!panduan(res[i]))
			{
				ans=0;
				break;
			}
		}}
	}

	if(ans==1)printf("YES
");
	else printf("NO
");

	return 0;
}