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 <cstdlib>
#include <algorithm>
#include <vector>
#include <cstring>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <cmath>
#define SIZE 500005

using namespace std;
typedef long long int ll;
typedef pair <int,int> P;

char str[SIZE];
int dp[SIZE];
int nxt[SIZE];

int main()
{
	int n;
	scanf("%d",&n);
	scanf("%s",&str);
	for(int i=0;i<=n;i++) dp[i]=n;
	int all=0;
	for(int i=n-1;i>=0;i--)
	{
		if(str[i]=='1') all++;
		nxt[i]=n;
		if(str[i]=='0') nxt[i]=i;
		else if(i+1<n) nxt[i]=nxt[i+1];
	}
	dp[0]=0;
	for(int i=0;i<n;i++)
	{
		if(str[i]=='0') dp[i+1]=min(dp[i],dp[i+1]);
		else
		{
			dp[i+1]=min(dp[i+1],dp[i]+1);
			if(nxt[i]==i+1)
			{
				for(int j=i+2;j<n;j++)
				{
					if(str[j]=='0') break;
					dp[j+1]=min(dp[j+1],dp[i]+1);
				}
			}
			else
			{
				if(nxt[i]+1<n&&str[nxt[i]+1]=='1')
				{
					dp[nxt[i]+2]=min(dp[nxt[i]+2],dp[i]+1);
				}
			}
		}
	}
	printf("%d\n",all-dp[n]);
	return 0;
}
