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;
const int maxn=500005;
int n, pre;
char s[maxn];
int dp[maxn];

int main()
{
    while(~scanf("%d%s", &n, s+1)){
        memset(dp, 0, sizeof(dp));
        for(int i=3;i<=n;i++){
            dp[i]=dp[i-1];
            if(s[i]=='0')pre=0;
            if(s[i]=='1'&&s[i-1]=='0'&&s[i-2]=='1'){
                int j=i-2;
                while(s[j]=='1'&&j){
                    dp[i]=max(dp[i], dp[j-1]+i-j-1);    //当前101作为末端
                    j--;
                }
                pre=i;
            }
            if(s[i]=='1'&&pre){
                dp[i]=max(dp[i], dp[pre-3]+i-pre+1);    //上一个101作为前端
            }
        }
        printf("%d\n", dp[n]);
    }
	return 0;
}
