Victor tries to write his own text editor, with word correction included. However, the rules of word correction are really strange.

Victor thinks that if a word contains two consecutive vowels, then it's kinda weird and it needs to be replaced. So the word corrector works in such a way: as long as there are two consecutive vowels in the word, it deletes the first vowel in a word such that there isanother vowel right before it. If there are no two consecutive vowels in the word, it is considered to be correct.

You are given a word s. Can you predict what will it become after correction?

In this problem letters a, e, i, o, u and y are considered to be vowels.

The first line contains one integer n (1 ≤ n ≤ 100) — the number of letters in word s before the correction.

The second line contains a string s consisting of exactly n lowercase Latin letters — the word before the correction.

Output the word s after the correction.

Explanations of the examples:

1. There is only one replace: weird  werd;
2. No replace needed since there are no two consecutive vowels;
3. aaeaa  aeaa  aaa  aa  a.

题目大意：给一个字符串，如果有两个连续的元音字母，则删掉后一个，一直进行这种操作，求最后的字符串.

分析：一个一个去枚举删就比较麻烦了，一个比较好的做法是用栈维护，枚举第i个位置，与栈顶的字符比较，看是否符合要求.最后输出栈里的字符串就好了.

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#include <cmath>

using namespace std;

typedef long long LL;

char s[],ans[];

int tot,n;

bool check(char a,char b)

{

    bool flag1 = false,flag2 = false;

    if (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u' || a == 'y')

        flag1 = true;

    if (b == 'a' || b == 'e' || b == 'i' || b == 'o' || b == 'u' || b == 'y')

        flag2 = true;

    if (flag1 && flag2)

        return true;

    return false;

}

int main()

{

    scanf("%d",&n);

    scanf("%s",s + );

    for (int i = ; i <= n; i++)

    {

        if (!tot)

            ans[++tot] = s[i];

        else

        {

            if (!check(ans[tot],s[i]))

                ans[++tot] = s[i];

        }

    }

    for (int i = ; i <= tot; i++)

        printf("%c",ans[i]);

    return ;

}