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;

const int MAXN = 5e5+5;
string s;

int n, l[MAXN];
int dp[MAXN];

int main() {
	cin >> n >> s;
	for (int i = 0; i < n; ++i) {
		if (i == 0) {
			l[i] = (s[i] == '1');
		} else {
			if (s[i] == '0')
				l[i] = 0;
			else
				l[i] = l[i-1]+1;
		}
	}

	dp[0] = dp[1] = dp[2] = 0;

	for (int i = 2; i < n; ++i) {
		//case 1: you have 101111....1
		dp[i+1] = dp[i];
		int k = l[i];
		if (k <= i-1 && s[i-k] == '0' && s[i-k-1] == '1')
			dp[i+1] = max(dp[i+1], dp[i-k-1]+k);
		if (s[i] == '1' && s[i-1] == '0' && s[i-2] == '1') {
			k = l[i-2];
			dp[i+1] = max(dp[i+1], dp[i-k-1]+k);
			if (k >= 2)
				dp[i+1] = max(dp[i+1], dp[i-k]+k-1);
		}
	}

	cout << dp[n] << endl;
}