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<algorithm>
#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
typedef long long LL;
const int N=500005;
int n; 
int pos[N];
char ss[N];
int f[N];
int main()
{
	scanf("%d",&n);
	scanf("%s",ss+1);
	for (int u=1;u<=n;u++)
	{
		if (ss[u]=='1')	pos[u]=pos[u-1];
		else pos[u]=u;
	}
	for (int u=1;u<=n;u++)
	{
		f[u]=f[u-1];
		if (ss[u]=='1')
		{
			if (ss[u-1]=='0'&&ss[u-2]=='1')
			{
				f[u]=max(f[u],f[pos[u-2]]+(u-2)-pos[u-2]);
				f[u]=max(f[u],f[pos[u-2]+1]+(u-2)-(pos[u-2]+1));
			}
			else if (pos[u]>1&&ss[pos[u]-1]=='1') f[u]=max(f[u],f[pos[u]-2]+u-pos[u]);
		}
	}
	printf("%d\n",f[n]);
	return 0;
}