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 = 500003;
int n;char s[MaxN];
int last[MaxN],dp[MaxN];
int main() {
    scanf("%d%s",&n,s+1);
    for (int i=1,p=0;i<=n;++i) {
        last[i]=p;
        if (s[i]=='0') p=i;
    }
    for (int i=1;i<=n;++i) {
        dp[i]=dp[i-1];
        if (s[i]=='0') continue;
        if (last[i]==0 || s[last[i]-1]=='0') continue;
        if (last[i]-2>=0) dp[i]=max(dp[i],dp[last[i]-2]+i-last[i]);
        if (last[i]==i-1) {
            int llast=last[last[i]];
            if (llast>=0) dp[i]=max(dp[i],dp[llast]+i-llast-2);
            if (llast+1<last[i]) dp[i]=max(dp[i],dp[llast+1]+i-llast-3);
        }
    }
    printf("%d\n",dp[n]);
    return 0;
}