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 LL long long
#define mst(a, b)	memset(a, b, sizeof a)
#define pill pair<int, int>
#define mk make_pair
const int qq = 5e5 + 10;
int n;
char st[qq];
int dp[qq];

int main() {
	scanf("%d%s", &n, st + 1);
	int f = -1;
	for(int i = 3; i <= n; ++i) {
		dp[i] = dp[i - 1];
		if(st[i] == '1'&& st[i - 1] == '0' && st[i - 2] == '1') {
			f = i;
			for(int j = i - 2; st[j] == '1' && j >= 1; --j) {
				dp[i] = max(dp[i], dp[j - 1] + i - j - 1);
			}
		}
		if(st[i] == '0')	f = -1;
		if(st[i] == '1' && f != -1) {
			dp[i] = max(dp[i], dp[f - 3] + i - f + 1);
		}
	}
	printf("%d\n", dp[n]);
}