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<iostream>
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cassert>
#include<cstring>
#include<climits>
#include<sstream>
#include<deque>
#include<vector>
#include<algorithm>
#include<set>
#include<map>
#include<bitset>

#define REP(i,s,n) for(int i=s;i<n;i++)
#define rep(i,n) REP(i,0,n)

using namespace std;

typedef long long ll;
const int IINF = INT_MAX;
const ll LLINF = LLONG_MAX;

int main(){
  int n;
  cin >> n;
  string s;
  cin >> s;
  int cnt = 0,i;
  for(i=0;i<n;++i) {
    if( s[i] == 'x' ) {
      ++cnt;
      if( cnt == 2 ) break;
    } else {
      cnt = 0;
    }
  }
  cout << i << endl;
  return 0;
}