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;
#define maxn 500005
int n,last[maxn],dp[maxn];
char s[maxn];
int main()
{	
	scanf("%d%s",&n,s+1);
	int lp=0;
	for (int i=1;i<=n;i++)
	{
		last[i]=lp;
		if (s[i]=='0') lp=i;
	}
	for (int i=1;i<=n;i++)
	{
		dp[i]=dp[i-1];
		if (s[i]=='0' || !last[i] || s[last[i]-1]=='0') continue;
		int pos=last[i];
		int lpos=last[pos];
		if (pos>=2) dp[i]=max(dp[i],dp[pos-2]+i-pos);//10111111...
		if (pos==i-1)//0111111...01
		{
			if (lpos>=0) dp[i]=max(dp[i],dp[lpos]+i-lpos-2);//dp[lpos]+number of 1 on the left
		if (lpos+1<pos) dp[i]=max(dp[i],dp[lpos+1]+i-lpos-3);
		}
	}
	printf("%d\n",dp[n]);
	return 0;
}
