C. Fixing Typos----栈

C. Fixing Typos

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Many modern text editors automatically check the spelling of the user's text. Some editors even suggest how to correct typos.

In this problem your task to implement a small functionality to correct two types of typos in a word. We will assume that three identical letters together is a typo (for example, word "helllo" contains a typo). Besides, a couple of identical letters immediately followed by another couple of identical letters is a typo too (for example, words "helloo" and "wwaatt" contain typos).

Write a code that deletes the minimum number of letters from a word, correcting described typos in the word. You are allowed to delete letters from both ends and from the middle of the word.

Input

The single line of the input contains word s, its length is from 1 to 200000 characters. The given word s consists of lowercase English letters.

Output

Print such word t that it doesn't contain any typos described in the problem statement and is obtained from s by deleting the least number of letters.

If there are multiple solutions, print any of them.

Examples

input

helloo

output

hello

input

woooooow

output

woow

Note

The second valid answer to the test from the statement is "heloo".

题目链接：http://codeforces.com/contest/363/problem/C

题目的意思是说给你一个字符串，现在需满足以下两个条件：

1.一个串若结尾的两个字符相同，则这个串后面的两个字符不能相同。

2.不存在连续相同的三个字符。

求最小的删除字符数。

我们可以用栈来模拟这个过程，按题意来说，有两种情况无法入栈

1. 如果当前的字母和栈顶的栈顶的字母相同，且栈顶第二个字母和栈顶第三个字母相同。
2. 如果当前字母和栈顶字母相同，且栈顶第二个字母和栈顶相同。

最后将栈内的元素从头到尾输出就好了。

代码：

#include <cstdio>#include <iostream>#include <cstring>using namespace std;char st[200010];char str[200010];int main(){    char ch;    scanf("%s",str);    int top=0;    int len=strlen(str);    st[top++]=str[0];    st[top++]=str[1];    for(int i=2;i<len;i++){        ch=str[i];        if(ch==st[top-1]&&st[top-2]==st[top-3]){            continue;        }        else if(ch==st[top-1]&&st[top-1]==st[top-2]){            continue;        }        else{            st[top]=ch;            top++;        }    }    st[top]='\0';    printf("%s\n",st);    return 0;}