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>
#define N 500005
using namespace std;
char str[N];
int n;
int l[N];
int dp[N];
void Init()
{
	scanf("%d",&n);
	scanf("%s",str+1);
	l[0]=0;
	for(int i=1;i<=n;i++)
	{
		if(str[i]=='1')l[i]=l[i-1];
		else l[i]=i;
	}
}
void Solve()
{
	dp[0]=0;
	for(int i=1;i<=n;i++)
	{
		dp[i]=dp[i-1];
		if(str[i]=='1')
		{
			if(str[i-1]=='0' and str[i-2]=='1')
			{
				dp[i]=max(dp[i],dp[l[i-2]]+(i-2)-l[i-2]);
				dp[i]=max(dp[i],dp[l[i-2]+1]+(i-2)-(l[i-2]+1));
			}
			else if(l[i]>1 and str[l[i]-1]=='1')
			{
				dp[i]=max(dp[i],dp[l[i]-2]+i-l[i]);
			}
		}
	}
}
void Print()
{
	printf("%d\n",dp[n]);
}
int main()
{
    Init();
    Solve();
	Print();
	//printf("%d\n",dp[-1]);
	return 0;
}