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,nex[501000],f[501000],pos;
char s[1001000];
void U(int &x,int y){
	x=max(x,y);
}
int main(){
	scanf("%d%s",&n,s+1); n=strlen(s+1);
	for (int i=n+1;i>0;i--)
		nex[i]= s[i]=='1'? nex[i+1]: i;
	for (int i=0;i+3<=n;i++){
		if (s[i+1]=='1'&&s[i+2]=='0'&&s[i+3]=='1'){
			pos=nex[i+3];
			for (int j=i+3;j<pos;j++)
				U(f[j],f[i]+j-i-2);
		}
		if (s[i+1]=='1'){
			pos=nex[i+1];
			if (pos+1<=n){
				if (s[pos+1]=='1'){
					U(f[pos+1],f[i]+pos-i-1);
				}
			}
		}
		U(f[i+1],f[i]);
	}
	for (int i=1;i<n;i++) U(f[n],f[i]);
	cout<<f[n];
}