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<cstdio>
#define maxn 500010
using namespace std;
char s[maxn];
int n;
int place[maxn],dp[maxn];
int main()
{
	scanf("%d",&n);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++)
	{
		if(s[i]=='0')
			place[i]=i;
		else
			place[i]=place[i-1];
	}
	for(int i=1;i<=n;i++)
	{
		dp[i]=dp[i-1];
		if(s[i]=='1')
		{
			if(s[i-1]=='0'&&s[i-2]=='1')
			{
				dp[i]=max(dp[i],dp[place[i-2]]+(i-2)-place[i-2]);
				dp[i]=max(dp[i],dp[place[i-2]+1]+(i-2)-place[i-2]-1);
			}
			else if(place[i]>1&&s[place[i]-1]=='1')
				dp[i]=max(dp[i],dp[place[i]-2]+i-place[i]);
		}
	}
	printf("%d\n", dp[n]);
	return 0;
}