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> pairInt;
#define FOR(i, n) for (int i = 0; i < int(n); i++)
#define FOR1(i, m, n) for (int i = int(m); i < int(n); i++)
#define MAX(a, b) ((a) >= (b) ? (a) : (b))
#define MIN(a, b) ((a) <= (b) ? (a) : (b))

int main(int argc, char *argv[])
{
  int N;

  scanf("%d", &N);

  bool flag = false;
  int i = 0;
  while (i < N) {
    char c;
    scanf(" %c", &c);
    if (flag == true && c == 'x') {
      break;
    } else if (flag == false && c == 'x') {
      flag = true;
    } else {
      flag = false;
    }
    i++;
  }

  printf("%d\n", i);
  
  return 0;
}