Codeforces 938.A Word Correction

A. Word Correction

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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

Output the word s after the correction.

Examples

input

Copy

5
weird

output

werd

input

Copy

4
word

output

word

input

Copy

5
aaeaa

output

a

Note

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[110],ans[110];
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 + 1);
    for (int i = 1; i <= n; i++)
    {
        if (!tot)
            ans[++tot] = s[i];
        else
        {
            if (!check(ans[tot],s[i]))
                ans[++tot] = s[i];
        }
    }
    for (int i = 1; i <= tot; i++)
        printf("%c",ans[i]);

    return 0;
}