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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

int dp[500010];

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
#ifdef LOCAL
	std::ifstream in("in");
	std::cin.rdbuf(in.rdbuf());
#endif

	string s;
	int N;
	cin >> N >> s;
	int last = 0;
	for(int i = 2; i < N; i++){
		dp[i] = dp[i - 1];
		if(s[i - 2] == '1' && s[i - 1] == '0' && s[i] == '1'){
			last = i;
			for(int j = i - 2; j >= 0; j--){
				if(s[j] == '0') break;
				dp[i] = max(dp[i], dp[j - 1] + i - j - 1);
			}
		}
		else if(s[i] == '0'){
			last = 0;
		}
		if(last){
			dp[i] = max(dp[i], dp[last - 3] + i - last + 1);
		}
	}
	cout << dp[N - 1] << endl;
}
