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 <climits>
#include <algorithm>
using namespace std;

int main() {
  int N; scanf("%d", &N);
  static char S[500003];
  scanf("%500000s", S);
  // NFA for (.|1+01|101+)*
  // q0 -0-> q0 (+0)
  // q0 -1-> q0 (+0)
  // q0 -1-> q1 (+1)
  // q1 -1-> q1 (+1)
  // q1 -0-> q2 (+0)
  // q2 -1-> q0 (+0)
  // q0 -1-> q3 (+0)
  // q3 -0-> q4 (+0)
  // q4 -1-> q4 (+1)
  // q4 -1-> q0 (+1)
  int dp[5] = {0, INT_MIN, INT_MIN, INT_MIN, INT_MIN};
  for(int i = 0; i < N; ++i) {
    int dp2[5] = {dp[0], dp[1], dp[2], dp[3], dp[4]};
    if(S[i] == '0') {
      // q0 -0-> q0 (+0)
      // q1 -0-> q2 (+0)
      // q3 -0-> q4 (+0)
      dp[0] = dp2[0];
      dp[1] = INT_MIN;
      dp[2] = dp2[1];
      dp[3] = INT_MIN;
      dp[4] = dp2[3];
    } else {
      // q0 -1-> q0 (+0)
      // q2 -1-> q0 (+0)
      // q4 -1-> q0 (+1)
      // q0 -1-> q1 (+1)
      // q1 -1-> q1 (+1)
      // q0 -1-> q3 (+0)
      // q4 -1-> q4 (+1)
      dp[0] = max({dp2[0], dp2[2], dp2[4] + 1});
      dp[1] = max({dp2[0] + 1, dp2[1] + 1});
      dp[2] = INT_MIN;
      dp[3] = dp2[0];
      dp[4] = dp2[4] + 1;
    }
  }
  printf("%d\n", dp[0]);
  return 0;
}
