problem

AOR Ika is studying to pass the test.

AOR Ika-chan solved the $ N $ question. After that, round the solved problem according to the following procedure.

1. Check the correctness of the answer.
2. If the answer is correct, write a circle mark, and if it is incorrect, write a cross mark on the answer sheet.



AOR Ika faints because of the fear of failing the test the moment she finds that the answer is wrong for $ 2 $ in a row. And no further rounding is possible.

Syncope occurs between steps $ 1 $ and $ 2 $.

You will be given an integer $ N $, which represents the number of questions AOR Ika has solved, and a string $ S $, which is a length $ N $ and represents the correctness of the answer. The string consists of'o'and'x', with'o' indicating the correct answer and'x' indicating the incorrect answer. The $ i $ letter indicates the correctness of the $ i $ question, and AOR Ika-chan rounds the $ 1 $ question in order.

Please output the number of questions that AOR Ika-chan can write the correctness.



output

Output the number of questions that AOR Ika-chan could write in the $ 1 $ line. Also, output a line break at the end.

Example

Input

3
oxx


Output

2
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;

int main() {

	int N;
	scanf("%d",&N);

	char* buf = new char[N+1];

	scanf("%s",buf);

	int ans = 0;

	char pre = '@';

	for(int i = 0; buf[i] != '\0'; i++){
		if(buf[i] == 'o'){
			ans++;
		}else if(buf[i] == 'x' && pre != 'x'){
			ans++;
		}else{
			break;
		}
		pre = buf[i];
	}

	printf("%d\n",ans);

	return 0;
}