Problem:
Score : 100 points

Problem Statement

Takahashi wants to print a document with N pages double-sided, where two pages of data can be printed on one sheet of paper.

At least how many sheets of paper does he need?

Constraints

N is an integer.

1 \leq N \leq 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

5

Sample Output 1

3

By printing the 1-st, 2-nd pages on the 1-st sheet, 3-rd and 4-th pages on the 2-nd sheet, and 5-th page on the 3-rd sheet, we can print all the data on 3 sheets of paper.

Sample Input 2

2

Sample Output 2

1

Sample Input 3

100

Sample Output 3

50

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

#include <bits/stdc++.h>
using namespace std;

#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define REP(i, n)  for(int i=0;i<(n);++i)
#define REPr(i, n) for(int i=(n)-1;i>=0; --i)
#define FORq(i, m, n) for(int i = (m);i <= (n);++i)
#define FORqr(i, m, n) for(int i = (n);i >=(m);--i)
#define PB push_back
#define MP make_pair
#define fst first
#define snd second
#define SIN(x, S) (S.count(x) != 0)
#define M0(x) memset(x,0,sizeof(x))
#define FILL(x, y) memset(x,y,sizeof(x))
#define MM(x) memset(x,-1,sizeof(x))
#define ALL(x) (x).begin(),(x).end()
#define DB(x) cerr << #x << " = " << x << endl
#define DB2(x, y) cerr << "(" << #x << ", " << #y << ") = (" << x << ", " << y << ")\n";
#define DEBUG int x12345; cin >> x12345;
typedef pair<int, int> PII;
typedef pair<long long, long long> PLL;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<long long> VL;
typedef long long ll;
typedef long long integer;
///////////////////////////////////////////////
const ll MOD = 1e9 + 7;
ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
 
template<class T>
inline bool chmin(T &a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
 
template<class T>
inline bool chmax(T &a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}
 
template<int MOD>
struct Fp {
	long long val;
 
	constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
		if (val < 0) v += MOD;
	}
 
	constexpr int getmod() { return MOD; }
 
	constexpr Fp operator-() const noexcept {
		return val ? MOD - val : 0;
	}
 
	constexpr Fp operator+(const Fp &r) const noexcept { return Fp(*this) += r; }
 
	constexpr Fp operator-(const Fp &r) const noexcept { return Fp(*this) -= r; }
 
	constexpr Fp operator*(const Fp &r) const noexcept { return Fp(*this) *= r; }
 
	constexpr Fp operator/(const Fp &r) const noexcept { return Fp(*this) /= r; }
 
	constexpr Fp &operator+=(const Fp &r) noexcept {
		val += r.val;
		if (val >= MOD) val = val % MOD;
		return *this;
	}
 
	constexpr Fp &operator-=(const Fp &r) noexcept {
		val -= r.val;
		if (val < 0) val += MOD * ((-val+MOD-1) / MOD);
		val = val % MOD;
		return *this;
	}
 
	constexpr Fp &operator*=(const Fp &r) noexcept {
		val = val * r.val % MOD;
		return *this;
	}
 
	constexpr Fp &operator/=(const Fp &r) noexcept {
		long long a = r.val, b = MOD, u = 1, v = 0;
		while (b) {
			long long t = a / b;
			a -= t * b;
			swap(a, b);
			u -= t * v;
			swap(u, v);
		}
		val = val * u % MOD;
		if (val < 0) val += MOD;
		return *this;
	}
 
	constexpr bool operator==(const Fp &r) const noexcept {
		return this->val == r.val;
	}
 
	constexpr bool operator!=(const Fp &r) const noexcept {
		return this->val != r.val;
	}
 
	friend constexpr ostream &operator<<(ostream &os, const Fp<MOD> &x) noexcept {
		return os << x.val;
	}
 
	friend constexpr istream &operator>>(istream &is, Fp<MOD> &x) noexcept {
		return is >> x.val;
	}
 
	friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
		if (n == 0) return 1;
		auto t = modpow(a, n / 2);
		t = t * t;
		if (n & 1) t = t * a;
		return t;
	}
};
 
using mint = Fp<MOD>;

///////////////////////////////////////////////
///  ((((っ・ω・)っ |AC|
int main(){
	int N; cin >> N;
	cout << (N + 1) / 2 << endl;
}