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<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

int main(){
    int n;
    cin>>n;
    string s;
    cin>>s;
    s+='x';
    int ans=0;
    rep(i,n){
      ++ans;
      if(s[i]=='x'&&s[i+1]=='x'){
        cout<<ans<<endl;
        return 0;
      }
    }
    cout<<ans<<endl;
    return 0;
}
