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 <iostream>
#include <string>
#include <algorithm>
using namespace std;
#pragma warning (disable: 4996)

string S; int pre[500009], n, dp[500009];

int main() {
	cin >> n >> S;
	pre[n] = n;
	for (int i = n - 1; i >= 0; i--) {
		if (S[i] == '1')pre[i] = pre[i + 1];
		else pre[i] = i;
	}
	for (int i = 0; i < n - 2; i++) {
		dp[i + 1] = max(dp[i + 1], dp[i]);
		if (S[i] == '0') continue;
		if (S[i + 1] == '1') {
			int to = pre[i];
			if (to + 1 >= S.size() || S[to + 1] == '0') continue;
			dp[to + 2] = max(dp[to + 2], dp[i] + (to - i));
		}
		else {
			if (S[i + 2] == '0') continue;
			dp[i + 3] = max(dp[i + 3], dp[i] + 1);
			int to = pre[i + 2];
			dp[to] = max(dp[to], dp[i] + (to - i - 2));
			if (to - i - 3 >= 1) dp[to - 1] = max(dp[to - 1], dp[i] + (to - i - 3));
		}
	}
	int ret = 0;
	for (int i = 0; i <= n; i++) ret = max(ret, dp[i]);
	cout << ret << endl;
	return 0;
}