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 <bits/stdc++.h>
using namespace std;
#define int long long
using ll=long long;
using vi=vector<int>;
using pii=pair<int,int>;
#define ALL(c) begin(c),end(c)
#define RALL(c) rbegin(c),rend(c)
#define ITR(i,b,e) for(auto i=(b);i!=(e);++i)
#define FORE(x,c) for(auto &x:c)
#define REPF(i,a,n) for(int i=a,i##len=(int)(n);i<i##len;++i)
#define REP(i,n) REPF(i,0,n)
#define REPR(i,n) for(int i=(int)(n);i>=0;--i)
#define SZ(c) ((int)c.size())
#define CONTAIN(c,x) (c.find(x)!=end(c))
#define OUTOFRANGE(y,x,h,w) (y<0||x<0||y>=h||x>=w)
#define dump(...)
const int DX[9]={0,1,0,-1,1,1,-1,-1,0},DY[9]={-1,0,1,0,-1,1,1,-1,0};
#define INF (1001001001)
#define INFLL (1001001001001001001ll)
template<class T> ostream& operator << (ostream &os,const vector<T> &v) {
    ITR(i,begin(v),end(v)) os<<*i<<(i==end(v)-1?"":"\n"); return os; }
template<class T> istream& operator >> (istream &is,vector<T> &v) {
    ITR(i,begin(v),end(v)) is>>*i; return is; }
template<class T> istream& operator >> (istream &is, pair<T,T> &p) {
    is>>p.first>>p.second; return is; }
template<class T> bool chmax(T &a,const T &b) {if(a<b) {a=b;return 1;} return 0;}
template<class T> bool chmin(T &a,const T &b) {if(b<a) {a=b;return 1;} return 0;}
template <class T> using heap = priority_queue<T,vector<T>,greater<T>>;
//------------------------------------------------------------------------------
struct before_main_function {
    before_main_function() {
        #ifdef int
            #undef INF
            #define INF INFLL
            #define stoi stoll
        #endif
        cin.tie(0);ios::sync_with_stdio(false);
        cout<<setprecision(15)<<fixed;
        #define endl "\n"
    }
} before_main_function;
//------------------8<------------------------------------8<--------------------

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