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
/******************************************
*    AUTHOR:         CHIRAG AGARWAL       *
*    INSTITUITION:   BITS PILANI, PILANI  *
******************************************/
#include <bits/stdc++.h>
using namespace std;
 
typedef long long LL; 
typedef long double LD;
const int MAX=5e5+5;
int pr[MAX];
int nxt[MAX];
int dp[MAX];


int main() 
{
	int n;
	scanf("%d",&n);
	string s;
	cin>>s;
	s=' '+s;
	for(int i=1;i<=n;i++)
	{
		if(s[i]=='0')
		{
			pr[i]=i;
		}
		else
		{
			pr[i]=pr[i-1];
		}
	}
	nxt[n+1]=n+1;
	for(int i=n;i>=1;i--)
	{
		if(s[i]=='0')
		{
			nxt[i]=i;
		}
		else
		{
			nxt[i]=nxt[i+1];
		}
	}

	for(int i=1;i<=n;i++)
	{
		if(i>=3&&s[i]=='1'&&s[i-1]=='0'&&s[i-2]=='1')
		{
			int lstone=pr[i-2]+1;
			dp[i]=max(dp[lstone-1]+(i-lstone-1),dp[i]);
			if(lstone!=(i-2))
			dp[i]=max(dp[lstone]+(i-lstone-2),dp[i]);
			lstone=nxt[i]-1;
			dp[lstone]=max(dp[lstone],dp[i-3]+(lstone-i+1));
			if(lstone!=i)
			dp[lstone-1]=max(dp[lstone-1],dp[i-3]+(lstone-i));
		}
		dp[i+1]=max(dp[i+1],dp[i]);
	}
	printf("%d\n",dp[n+1]);
	
	return 0;
}
