Codeforces Round-#380-Div. 2-A-Interview with Oleg

Codeforces Round #380 (Div. 2, Rated, Based on Technocup 2017 - Elimination Round 2)

A. Interview with Oleg

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Polycarp has interviewed Oleg and has written the interview down without punctuation marks and spaces to save time. Thus, the interview is now a string s consisting of n lowercase English letters.

There is a filler word ogo in Oleg’s speech. All words that can be obtained from ogo by adding go several times to the end of it are also considered to be fillers. For example, the words ogo, ogogo, ogogogo are fillers, but the words go, og, ogog, ogogog and oggo are not fillers.

The fillers have maximal size, for example, for ogogoo speech we can’t consider ogo a filler and goo as a normal phrase. We should consider ogogo as a filler here.

To print the interview, Polycarp has to replace each of the fillers with three asterisks. Note that a filler word is replaced with exactly three asterisks regardless of its length.

Polycarp has dealt with this problem in no time. Can you do the same? The clock is ticking!

Input

The first line contains a positive integer n (1 ≤ n ≤ 100) — the length of the interview.

The second line contains the string s of length n, consisting of lowercase English letters.

Output

Print the interview text after the replacement of each of the fillers with “* * “. It is allowed for the substring “**” to have several consecutive occurences.

Examples

input

7

aogogob

output

a***b

input

13

ogogmgogogogo

output

gmg

input

9

ogoogoogo

output

Note

The first sample contains one filler word ogogo, so the interview for printing is “a* **b”.

The second sample contains two fillers ogo and ogogogo. Thus, the interview is transformed to ” * * * gmg * * *”.

题意：对于给定字符串，如果遇到ogo则输出三个星号，如果ogo后面跟着几个go那么忽略这些go，遇到其他组合的字符原样输出即可。

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
const int maxn=105;
char str[maxn];
int main()
{
memset(str,'\0',sizeof(str));
int len;
scanf("%d%s",&len,str);
for(int i=0;i<len;)
{
if(str[i]=='o'&&str[i+1]=='g'&&str[i+2]=='o')
{
printf("***");
i=i+3;
while(str[i]=='g'&&str[i+1]=='o')
i=i+2;
}
else
{
printf("%c",str[i]);
i++;
}
}
printf("\n");
}