N cells are arranged in a row. Some of them may contain tokens. You are given a string s that consists of `0`s and `1`s. If the i-th character of s is `1`, the i-th cell (from left) contains a token. Otherwise, it doesn't contain a token.

Snuke wants to perform the following operation as many times as possible. In each operation, he chooses three consecutive cells. Let's call the cells X, Y, Z from left to right. In order for the operation to be valid, both X and Z must contain tokens and Y must not contain a token. Then, he removes these two tokens and puts a new token on Y.

How many operations can he perform if he performs operations in the optimal way?

Constraints

* 1 \leq N \leq 500,000
* |s| = N
* Each character in s is either `0` or `1`.

Input

Input is given from Standard Input in the following format:


N
s


Output

Print the answer.

Examples

Input

7
1010101


Output

2


Input

50
10101000010011011110001001111110000101010111100110


Output

10
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

char buff[1<<20];
int n;

int solve(int start, int end)
{
  if (start == end) return 0;
  int a = 0, b = 0, c = 0, x = 0;
  for (int i = start; i <= end;)
  {
    int j = i;
    while (j <= end && buff[j] == '1') ++j;
    int cnt = j - i;
    i = j + 1;
    if (x > 0)
    {
      int aa = c + x;
      if (x >= 2) aa = max(aa, a + x - 1), aa = max(aa, b);
      int bb = c + cnt;
      if (x >= 2) bb = max(bb, a + cnt), bb = max(bb, b + cnt - 1);
      int cc = max(max(a, b), c);
      a = aa;
      b = bb;
      c = cc;
    }
    x = cnt;
  }
  return max(max(a, b), c);
}
int main()
{
  scanf("%d %s", &n, buff);
  buff[n] = '0';
  buff[n+1] = '0';
  buff[n+2] = 0;
  n += 2;
  int ans = 0;
  for (int curr = 0; curr < n;)
  {
    while (curr < n && buff[curr] == '0')
      ++curr;
    if (curr == n) break;
    int start = curr++;
    while (curr < n && (buff[curr] == '1' || buff[curr-1] == '1')) ++curr;
    ans += solve(start, curr - 2);
    ++curr;
  }
  printf("%d", ans);
  
  return 0;
}
/*

*/