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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> P;
typedef pair<ll, pair<ll, ll> > P3;

const ll MOD = ll(1e9 + 7);
const ll LLINF = LLONG_MAX;
const int IINF = INT_MAX;
const int MAX_N = int(1e5) + 5;
const double EPS = 1e-8;
const int di[] = {0, 1, 0, -1}, dj[] = {1, 0, -1, 0};
string dir[] = {"R", "D", "L", "U"};
#define REP(i, n) for (int i = 0; i < n; i++)
#define REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define ALL(v) (v).begin(), (v).end()

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

