Problem:
Score : 700 points

Problem Statement

N cells are arranged in a row.
Some of them may contain tokens.
You are given a string s that consists of 0s and 1s.
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.

Sample Input 1

7
1010101

Sample Output 1

2

For example, he can perform two operations in the following way:

Perform an operation on the last three cells. Now the string that represents tokens becomes 1010010.

Perform an operation on the first three cells. Now the string that represents tokens becomes 0100010.

Note that the choice of operations matters.
For example, if he chooses three cells in the middle first, he can perform no more operations.

Sample Input 2

50
10101000010011011110001001111110000101010111100110

Sample Output 2

10

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

////////////////////////////////////////
//
//
////////////////////////////////////////

//#define debug //*******************************************************************************************************************************************
#ifdef debug
#include <chrono>
#endif

#include <iostream>
#include <algorithm> // next_permutation
#include <iomanip>
#include <cmath>
#include <vector>
#include <sstream>
#include <string>
#include <cstring> //memcpy
#include <cstdio>
#include <stack>
#include <queue>
#include <list>
#include <numeric> //accumulate
//#include <unordered_map> //hash func.
#include <fstream> //ifstream, ofstream

//#define NDEBUG //If NDEBUG is defined before #include <cassert>, assert will be ignored. You had better define NDEBUG when u submit the code.
#include <cassert> //assert

using namespace std;


#define dout cout
//If u wanna output to a text file instead of standard output, plz define OUTPUTFILE.
//#define OUTPUTFILE "output.txt" //*******************************************************************************************************************************************
#ifdef OUTPUTFILE
#define dout outputfile //というのはTLE問題でどうせdoutを全部コメントアウトしなくちゃいけなくなるので上記方式は却下して、代わりに出力先を標準出力からファイルに切り替えできるようにする
ofstream outputfile(OUTPUTFILE);
#define OutputFilePath "/Users/Nag/Documents/Prgm/Test/DerivedData/Test/Build/Products/Debug/output.txt"
#endif


#define din cin
//If u wanna input from a text file instead of standard input, plz define INPUTFROMTEXTFILEを.
//#define INPUTFILE "input.txt" //*******************************************************************************************************************************************
#ifdef INPUTFILE
#define din inputfile
ifstream inputfile(INPUTFILE);
#endif

#define scan(A) scanf("%d", &(A))
#define disp(A) dout << #A << " = " << setw(3) << (A) << endl
#define disP(A) dout << setw(3) << (A) << " "
#define rep(i,a,n) for(int (i)=(a); (i)<(n); (i)++)
#define dispAll(A,n) dout << #A << " = "; rep(j, 0, (n)) {disP(A[j]);} dout << endl
//#define dispAll(A,n) cout << #A << " = "; rep(j, 0, (n)) {cout << setw(3) << A[j] << " ";} cout << endl

#define p(i) (i)/2
#define l(i) (i)*2
#define r(i) (i)*2+1

typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

const int INF = 2e9+10;
const ll INF_LL = 1e18; //Be careful for overflow.
const int NONE = -1;

#define N_MAX 500000 //num of vertex or element
#define M_MAX 100000 //num of edge

int N, M;
int s[N_MAX];

int main() {
    
    //cin, coutの高速化
    cin.tie(0); //cinとcoutの同期を切る
    ios::sync_with_stdio(false); //iostreamとstdioの同期を切る
    
    //read input data
    scan(N);
    char c;
    rep(i,0,N) {
        scanf(" %c",&c);
        s[i] = c - '0';
    }
    
#ifdef debug
    rep(i,0,N) {
        disP(s[i]);
    }
    dout << endl;
#endif

    
    
    
    //------------------------------------------------------------------------------------------
#ifdef debug
    //start timer
    auto startTime = chrono::system_clock::now();
#endif
    //------------------------------------------------------------------------------------------
    
    int ans = 0;
    
    int l = 0;
    int r = N-1;
    while(l<=r) {
        
        if( l+2<N && s[l]==1 && s[l+1]==0 && s[l+2]==1) {
            ans++;
            s[l]=0; s[l+1]=1; s[l+2]=0;
        }
        l++;
        
        if( r-2>=0 && s[r-2]==1 && s[r-1]==0 && s[r]==1) {
            ans++;
            s[r-2]=0; s[r-1]=1; s[r]=0;
            r += 2;
        }
        r--;
        
        
#ifdef debug
        rep(i,0,N) {
            if(i==l or i==r) {
                dout << " [" << s[i] << "]";
            }
            else {
                disP(s[i]);
            }
        }
        dout << endl;
#endif
    }

    printf("%d\n", ans);
    
    
    
    //------------------------------------------------------------------------------------------
#ifdef debug
    //stop timer
    auto endTime = chrono::system_clock::now();
    auto dur = endTime - startTime;
    auto msec = chrono::duration_cast<chrono::milliseconds>(dur).count();
    dout << fixed << setprecision(4) << (double)msec/1000 << " sec \n";
#endif
    //------------------------------------------------------------------------------------------
    
#ifdef INPUTFILE
    inputfile.close();
#endif
    
#ifdef OUTPUTFILE
    outputfile.close();
    cout << "\"" << OutputFilePath << "\"" << endl;
#endif
    
    return 0;
}