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
//2017-10-9
//miaomiao
//
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;

#define For(i, a, b) for(int i = (a); i <= (int)(b); ++i)
#define Forr(i, a, b) for(int i = (a); i >= (int)(b); --i)

#define N (500000+5)

char s[N];
int f[N];

int main(){
	int n;
	scanf("%d%s", &n, s+1);
	
	int lst = 0;

	For(i, 2, n){
		f[i] = f[i-1];

		if(s[i] == '1' && s[i-1] == '0' && s[i-2] == '1'){
			lst = i;
			Forr(j, i-2, 1){
				if(s[j] == '0') break;
				f[i] = max(f[i], f[j-1] + ((i-2)-j+1));
			}
		}

		if(s[i] == '0') lst = 0;	
		if(lst) f[i] = max(f[i], f[lst-3] + (i-lst+1));
		
	//	printf("f[%d] = %d\n", i, f[i]);
	}
	
	printf("%d\n", f[n]);

	return 0;
}