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"
#define REP(i, n) for(int i = 0; i < n; i++)
#define REPR(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define INF 2e9
#define ALL(v) v.begin(), v.end()
typedef long long ll;

using namespace std;

// init {{{
struct ProconInit {
    static constexpr int IOS_PREC = 15;
    static constexpr bool AUTOFLUSH = false;

    ProconInit() {
        cin.tie(nullptr);
        ios::sync_with_stdio(false);
        cout << fixed << setprecision(IOS_PREC);
        if (AUTOFLUSH)
            cout << unitbuf;
    }
} PROCON_INIT;


const int NM = 20000;//ここを問題の制約を見ながらいじる．配列の確保に使う．
ll A[NM];//あまり褒められたことではないが関数の外側に配列を確保すると怒られにくい．
bool Minus[NM];//あまり褒められたことではないが関数の外側に配列を確保すると怒られにくい．

         //ll dp[NM];//あまり褒められたことではないが関数の外側に配列を確保すると怒られにくい．
//string S[NM];//あまり褒められたことではないが関数の外側に配列を確保すると怒られにくい．
//int B[NM];//あまり褒められたことではないが関数の外側に配列を確保すると怒られにくい．
//double Perf[NM];

ll MOD = 1000000007;
int digit_length(int num)
{
    int count = 0;
    while (num) {
        num = num / 10;
        ++count;
    }
    return count;
}


ll gcd(ll a, ll b) {
    ll v0 = a, v1 = b, v2 = a % b;
    while (v2 != 0) {
        v0 = v1;
        v1 = v2;
        v2 = v0 % v1;
    };
    return v1;
}


// mod. m での a の逆元 a^{-1} を計算する
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

//ll lcm(ll a, ll b) {
//    return a / (gcd(a, b) % MOD) * (b % MOD);
//}

ll lcm(ll a, ll b) {
    return( (a % MOD) *modinv(gcd(a, b), MOD) * (b % MOD)) % MOD;
}
       // Main program


void solve() {
    ll  N;
    cin >> N;
 
    cout << (int)(((double)N/2.0)+0.5);

    return;
}



int main() {
    solve();
    return 0;
}