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 <cstdlib>
#include <vector>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>

#pragma comment(linker, "/STACK:336777216")  

using namespace std;

const int MAXN = 500000 + 10;

int N;
char S[MAXN];
int DP[MAXN];

void Work()
{
	scanf("%d%s", &N, S + 1);
	int lastone = 0;
	for (int i = 1; i <= 2; i ++)
	{
		if (S[i] == '1') lastone ++;
		else lastone = 0;
	}
	for (int i = 3; i <= N; i ++)
	{
		if (S[i] == '1') lastone ++;
		else lastone = 0;

		DP[i] = DP[i - 1];
		if (S[i] == '1' && S[i - 1] == '0')
		{
			for (int j = i - 2; j >= 1; j --)
				if (S[j] == '1')
					DP[i] = max(DP[i], DP[j - 1] + (i - 1 - j));
				else
					break;
		}
		
		if (lastone >= 1 && i - lastone - 1 >= 1 && S[i - lastone] == '0' && S[i - lastone - 1] == '1')
			DP[i] = max(DP[i], DP[i - lastone - 2] + lastone);
	}
	printf("%d\n", DP[N]);
}

int main()
{
	Work();
	return 0;
}