Problem:
Score : 100 points

Problem Statement

There are N students in a school.

We will divide these students into some groups, and in each group they will discuss some themes.

You think that groups consisting of two or less students cannot have an effective discussion, so you want to have as many groups consisting of three or more students as possible.

Divide the students so that the number of groups consisting of three or more students is maximized.

Constraints

1 \leq N \leq 1000

All input values are integers.

Input

Input is given from Standard Input in the following format:

N

Output

If you can form at most x groups consisting of three or more students, print x.

Sample Input 1

8

Sample Output 1

2

For example, you can form a group of three students and another of five students.

Sample Input 2

2

Sample Output 2

0

Sometimes you cannot form any group consisting of three or more students, regardless of how you divide the students.

Sample Input 3

9

Sample Output 3

3

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

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<string>
#include<sstream>
#include<iomanip>
#include<utility>
#include<cmath>
#include<set>
#include<list>
#include<string>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<set>
#include<cstring>
#include<iterator>
#include<bitset>
#include<assert.h>
//#include<chrono>
//#include<time.h>
//#include<windows.h>
using namespace std;
 
//#define int ll
typedef long long ll;
typedef pair<int,int> P;
typedef pair<int,pair<int,int> > PP;
typedef vector<int> VI;
typedef vector<long long int> VL;
static const int MOD = 1000000007;
//static const int INF = 2147483647;
//static const long long INF = 9223372000000000000;
//static const long long INF = 9223372000000000000/2;
//static const int INF = 1000010000;
//int dx4[4] = {0,1,0,-1}, dy4[4] = {-1,0,1,0};
//int dx5[5] = {-1,0,0,0,1}, dy5[5] = {0,-1,0,1,0};
//int dx8[8] = {-1,0,1,1,1,0,-1,-1}, dy8[8] = {1,1,1,0,-1,-1,-1,0};
//int dx9[9] = {-1,0,1,1,1,0,-1,-1,0}, dy9[9] = {1,1,1,0,-1,-1,-1,0,0};
#define PB push_back
#define MP make_pair
#define MT make_tuple
#define FI first
#define SE second
#define NP next_permutation
#define PQ priority_queue
#define UB upper_bound
#define LB lower_bound
#define SZ(a) int((a).size())
#define LEN(a) int((a).length())
#define SORT(c) sort((c).begin(),(c).end())
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,x) for(int i=0;i<(int)(x);i++)
#define REP1(i,x) for(int i=1;i<=(int)(x);i++)
#define RREP(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define RREP1(i,x) for(int i=((int)(x));i>0;i--)
#define ALL(x) (x).begin(),(x).end()
#define YESNO(x) puts((x)?"YES":"NO")
struct edge {int /*from,*/to,cost;};

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

    printf("%d\n",n/3);

    return 0;
}