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;
//type
typedef long long ll;
#define pii pair<int, int>
#define vi vector<int>
//x * y * 1.0 can cause overflow
//constant
#define inf (int)(1e9+7)
#define mod (ll)(1e9+7)
#define eps 1e-10
//omission
#define eb emplace_back
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define all(v) v.begin(), v.end()
#define ios cin.tie(0); ios::sync_with_stdio(false)
#define FIX(a) cout << fixed << setprecision(a)
//manip
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define UNIQUE(v) v.erase(unique(v.begin(), v.end(), v.end());
#define fill(x, y) memset(x, y, sizeof(x))
#define ceil(a, b) a / b + !!(a % b)
template<class T> T power(T a, T b)
{return b ? power(a * a % inf, b / 2) * (b % 2 ? a : 1) % inf : 1;}
#define LB(v, x) (int)(lower_bound(v.begin(), v.end(), x) - v.begin())
#define UB(v, x) (int)(upper_bound(v.begin(), v.end(), x) - v.begin())
template<typename T> T gcd(T a, T b) {if (b == 0) return a; return gcd(b, a % b);}
template<typename T> T lcm(T a, T b) {return a / gcd(max(a, b), min(a, b)) * b;} //先に割る!!
#define outa(a, n) rep(i, n) cout << a[i] << " "; cout << endl
#define out(a) cout << a << endl
//loop
#define rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define rep3(i, st, n) for (int i = st; i < n; ++i)
//algorithm
//double pointer, l start, how many adds, can be 0 -> init r = l, sum = 0
//bfs, not x, y, i, j
//not !(i % 2), i % 2 == 0
//minimal required terms
//cast caution
//look constraints always



int main() {
    int n; cin >> n;
    string s; cin >> s;
    int ans = 1;
    rep3(i, 1, SZ(s)) {
        if (s[i] == s[i - 1] && s[i] == 'x') {
            break;
        }
        ans++;
    }
    out(ans);
}

