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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

int main(){
	int n;
	string s; cin>>n>>s;

	vector<int> pos={-1};
	rep(i,n) if(s[i]=='0') pos.emplace_back(i);
	pos.emplace_back(n);

	vector<vector<int>> Ev(n); // right endpoint -> left endpoint
	for(int i=1;i<n-1;i++) if(s[i-1]=='1' && s[i]=='0' && s[i+1]=='1') {
		for(int x=i-1;x>=0&&s[x]=='1';x--) Ev[i+1].emplace_back(x);
		for(int x=i+2;x< n&&s[x]=='1';x++) Ev[x].emplace_back(i-1);
	}

	vector<int> dp(n+1);
	rep(x,n){
		dp[x+1]=dp[x];
		for(int l:Ev[x]){
			dp[x+1]=max(dp[x+1],dp[l]+x-l-1);
		}
	}
	cout<<dp[n]<<'\n';

	return 0;
}
