Codeforces Round #497 (Div. 2) A. Romaji

A. Romaji

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vitya has just started learning Berlanese language. It is known that Berlanese uses the Latin alphabet. Vowel letters are "a", "o", "u", "i", and "e". Other letters are consonant.

In Berlanese, there has to be a vowel after every consonant, but there can be any letter after any vowel. The only exception is a consonant "n"; after this letter, there can be any letter (not only a vowel) or there can be no letter at all. For example, the words "harakiri", "yupie", "man", and "nbo" are Berlanese while the words "horse", "king", "my", and "nz" are not.

Help Vitya find out if a word $s$ is Berlanese.

Input

The first line of the input contains the string $s$ consisting of $|s|$ ($1\le |s|\le 100$) lowercase Latin letters.

Output

Print "YES" (without quotes) if there is a vowel after every consonant except "n", otherwise print "NO".

You can print each letter in any case (upper or lower).

Examples

input

Copy

sumimasen

output

Copy

YES

input

Copy

ninja

output

Copy

YES

input

Copy

codeforces

output

Copy

NO

Note

In the first and second samples, a vowel goes after each consonant except "n", so the word is Berlanese.

In the third sample, the consonant "c" goes after the consonant "r", and the consonant "s" stands on the end, so the word is not Berlanese.

判断字符串中除字母‘n’外每个非元音字母后是否有元音字母（为空也不行）

没有错

全有对

‘n’后可跟任意字母 包括空字母

#include <iostream>
using namespace std;

int main()
{
	string a;
	cin>>a;
	
	for(int i = 0; i < a.length(); i++)
	{
		if(a[i] == 'n')
			continue;
			
		else if(a[i] != 'a' && a[i] != 'e' && a[i] != 'i' && a[i] != 'o' && a[i] != 'u')
		{
			if(a[i + 1] != 'a' && a[i + 1] != 'e' && a[i + 1] != 'i' && a[i + 1] != 'o' && a[i + 1] != 'u')
			{
				cout<<"NO"<<endl;
				goto l1;
			}
		}
	}
	cout<<"YES"<<endl;
	
	l1:
	return 0;
}