Problem:
Score : 300 points

Problem Statement

You are given an integer sequence of length n, a_1, ..., a_n.
Let us consider performing the following n operations on an empty sequence b.

The i-th operation is as follows:

Append a_i to the end of b.

Reverse the order of the elements in b.

Find the sequence b obtained after these n operations.

Constraints

1 \leq n \leq 2\times 10^5

0 \leq a_i \leq 10^9

n and a_i are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print n integers in a line with spaces in between.
The i-th integer should be b_i.

Sample Input 1

4
1 2 3 4

Sample Output 1

4 2 1 3

After step 1 of the first operation, b becomes: 1.

After step 2 of the first operation, b becomes: 1.

After step 1 of the second operation, b becomes: 1, 2.

After step 2 of the second operation, b becomes: 2, 1.

After step 1 of the third operation, b becomes: 2, 1, 3.

After step 2 of the third operation, b becomes: 3, 1, 2.

After step 1 of the fourth operation, b becomes: 3, 1, 2, 4.

After step 2 of the fourth operation, b becomes: 4, 2, 1, 3.

Thus, the answer is 4 2 1 3.

Sample Input 2

3
1 2 3

Sample Output 2

3 1 2

As shown above in Sample Output 1, b becomes 3, 1, 2 after step 2 of the third operation. Thus, the answer is 3 1 2.

Sample Input 3

1
1000000000

Sample Output 3

1000000000

Sample Input 4

6
0 6 7 6 7 0

Sample Output 4

0 6 6 0 7 7

Solution in C++11 (GCC 4.8.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <iomanip>
#include <vector>
#include <valarray>
#include <map>
#include <set>
#include <list>
#include <queue>
#include <stack>
#include <bitset>
#include <utility>
#include <numeric>
#include <algorithm>
#include <functional>
#include <complex>
#include <string>
#include <sstream>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <cmath>

using namespace std;

#define all(c) c.begin(),c.end()
#define repeat(i,n) for(int i=0;i<static_cast<int>(n);i++)
#define debug(x) #x << "=" << (x)
#define dump(x) cerr << debug(x) << " (L:" << __LINE__ << ")"<< endl

typedef long long ll;
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef vector<long> vl;
typedef vector<vector<long> > vvl;
typedef vector<string> vs;

template<typename T>
ostream& operator<<(ostream& os, const vector<T>& vec){
    os << "[";
    for(int i = 0; i < vec.size(); i++){
        os << vec[i] << ",";
    }
    os << "]";
    return os;
}

template<typename T>
T input(){
    T t;
    cin >> t;
    return t;
}

template<typename T>
vector<T> input(const int N){
    vector<T> v(N);
    repeat(i,N) cin >> v[i];
    return v;
}

long long gcd(long long a, long long b){
    if(b == 0){
        return a;
    }
    return gcd(b, a % b);
}

long long lcm(long long a, long long b){
    return (a / gcd(a, b)) * b;
}

long long mul(const long long& a, const long long& b, const long long& mod) {
    return ((a % mod) * (b % mod)) % mod;
}

long long power(const long long& x, const long long& y, const long long& mod) {
    if (y == 0) {
        return 1;
    } else if (y == 1) {
        return x % mod;
    } else {
        long long value = power(x, y / 2, mod);
        if (y % 2 == 0) {
            return mul(value, value, mod);
        } else {
            return mul(value, value, mod) * x % mod;
        }
    }
}

long long div(const long long& a, const long long& b, const long long& mod) {
    return mul(a, power(b, mod - 2, mod), mod);
}

map<long long, long long> factorials;
long long factorial(const long long& n, const long long& mod) {
    if (n == 0 || n == 1) {
        return 1;
    }
    if (factorials[n] != 0) {
        return factorials[n];
    }
    factorials[n] = n * factorial(n - 1, mod) % mod;
    return factorials[n] % mod;
}

long long combination(const long long& n, const long long& x, const long long& mod) {
    long long numerator = 1;
    long long denominator = 1;
    repeat(i, x) {
        numerator *= (n - i) % mod;
        numerator %= mod;
        denominator *= (i + 1) % mod;
        denominator %= mod;
    }
    return div(numerator, denominator, mod);
}

int main(){
    int N;
    cin >> N;

    long number;
    vl numbers;
    repeat (i, N) {
        cin >> number;
        if (i % 2 == 0) {
            numbers.push_back(number);
        } else {
            numbers.insert(numbers.begin(), number);
        }
    }
    if (N % 2 == 0) {
        cout << numbers[0];
        for (int i = 1; i < numbers.size(); ++i) {
            cout << " " << numbers[i];
        }
    } else {
        cout << numbers[N - 1];
        for (int i = numbers.size() - 2; 0 <= i; --i) {
            cout << " " << numbers[i];
        }
    }

    cout << endl;
    return 0;
}