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<functional>
#include<algorithm>
#include<iostream>
#include<utility>
#include<string>
#include<vector>
#include<cmath>
#include<stack>
#include<queue>
#include<deque>
#include<map>
#include<set>
#define rep(i,n) for(int i=0;i<n;i++)
#define loop(i,x,n) for(int i=x;i<n;i++)
#define pb push_back
#define int long long
using namespace std;
const int INF=1000000007;
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};

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