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>
using namespace std;
int n,dp[500005],nxt[500005];
char s[500005];
int main()
{
	scanf("%d",&n);
	scanf("%s",&s);
	for(int i=1;i<=n;i++)dp[i]=500005;
	int s1=0;
	for(int i=n-1;i>=0;i--)
	{
		if(s[i]=='1')s1++;
		if(s[i]=='0')nxt[i]=i;else if(i<n-1)nxt[i]=nxt[i+1];else nxt[i]=n;
	}
	dp[0]=0;
	for(int i=0;i<n;i++)
	{
		if(s[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&&s[j]=='1';j++)dp[j+1]=min(dp[j+1],dp[i]+1);
			else
				if(nxt[i]<n&&s[nxt[i]+1]=='1')dp[nxt[i]+2]=min(dp[nxt[i]+2],dp[i]+1);
		}
	}
	printf("%d\n",s1-dp[n]);
	return 0;
}