Problem Description

Password security is a tricky thing. Users prefer simple passwords that are easy to remember (like buddy), but such passwords are often insecure. Some sites use random computer-generated passwords (like xvtpzyo), but users have a hard time remembering them and sometimes leave them written on notes stuck to their computer. One potential solution is to generate "pronounceable" passwords that are relatively secure but still easy to remember.

FnordCom is developing such a password generator. You work in the quality control department, and it's your job to test the generator and make sure that the passwords are acceptable. To be acceptable, a password must satisfy these three rules:

It must contain at least one vowel.

It cannot contain three consecutive vowels or three consecutive consonants.

It cannot contain two consecutive occurrences of the same letter, except for 'ee' or 'oo'.

(For the purposes of this problem, the vowels are 'a', 'e', 'i', 'o', and 'u'; all other letters are consonants.) Note that these rules are not perfect; there are many common/pronounceable words that are not acceptable.

 

Input

The input consists of one or more potential passwords, one per line, followed by a line containing only the word 'end' that signals the end of the file. Each password is at least one and at most twenty letters long and consists only of lowercase letters.

 

Output

For each password, output whether or not it is acceptable, using the precise format shown in the example.

 

Sample Input

 

Sample Output

 

题意：给你一段密码，判断这段密码是否安全，若安全则是 acceptable。

密码必须同时满足下面三个限制条件才是安全密码：1、必须包含元音字母；

2、不能包含连续的三个元音或辅音字母；

3、除了ee和oo外，任何两个连续字母如果相同，密码就不安全。

#include<stdio.h>

#include<string.h>

int main()

{

    int i,flag,len,vowel;

    char str[25];

    while(gets(str))

    {

        if(!strcmp(str,"end"))  break;

        len=strlen(str);

        vowel=0; flag=1;

        for(i=0;i<=1;i++) /*特判str[0]和str[1]*/

           if(str[i]=='a'||str[i]=='e'||str[i]=='i'||str[i]=='o'||str[i]=='u')

             vowel++;

        if(str[0]==str[1]&&str[0]!='o'&&str[0]!='e')

        {

            flag=0;

            printf("<%s> is not acceptable.\n",str);

            continue;

        }

        for(i=2;i<len;i++)

        {

            if((str[i]==str[i-1])&&(str[i]!='e'&&str[i]!='o')) /*两个连续*/

            {

                flag=0;

                break;

            }

            if(str[i]=='a'||str[i]=='e'||str[i]=='i'||str[i]=='o'||str[i]=='u') /*元音字母*/

                vowel++;

            if((str[i]=='a'||str[i]=='e'||str[i]=='i'||str[i]=='o'||str[i]=='u')&&

              (str[i-1]=='a'||str[i-1]=='e'||str[i-1]=='i'||str[i-1]=='o'||str[i-1]=='u')&&

              (str[i-2]=='a'||str[i-2]=='e'||str[i-2]=='i'||str[i-2]=='o'||str[i-2]=='u'))   /*三个连续元音*/

            {

                flag=0;

                break;

            }

            if((str[i]!='a'&&str[i]!='e'&&str[i]!='i'&&str[i]!='o'&&str[i]!='u')&&

              (str[i-1]!='a'&&str[i-1]!='e'&&str[i-1]!='i'&&str[i-1]!='o'&&str[i-1]!='u')&&

              (str[i-2]!='a'&&str[i-2]!='e'&&str[i-2]!='i'&&str[i-2]!='o'&&str[i-2]!='u')) /*三个连续辅音*/

            {

                flag=0;

                break;

            }

        }

        if(flag&&vowel>0)

            printf("<%s> is acceptable.\n",str);

        else

            printf("<%s> is not acceptable.\n",str);

    }

    return 0;

}