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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <stdio.h>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <limits>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <iterator>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <fstream>
#include <queue>

#define MAX 100009
#define INF 99999999
#define in(a,b)  ( (b).find(a) != (b).end())
#define clr(a,b) memset((a), (b), sizeof((a)))
#define pb       push_back
#define all(a)   a.begin(), a.end()  
#define mp       make_pair
#define ll	long long
#define fi	first
#define se second

const long double pi = 3.141592653589793238L;
const ll mod = 1e9 + 7;

/**
*
* @Author Lahia omar
*/

using namespace std;
void split(vector<string> &v, string cur) {
	istringstream iss(cur);
	copy(istream_iterator<string>(iss), istream_iterator<string>(), back_inserter(v));
}
int n;
vector<ll> tab, outL,outR;
void ff() {
	for (int i = outR.size() - 1; i >= 0; i--) {
		cout << outR[i] << ' ';
	}
	for (int i = 0; i < outL.size(); i++) {
		cout << outL[i] << ' ';
	}
}
void f() {
	for (int i = outL.size() - 1 ; i >= 0; i--) {
		cout << outL[i] << ' ';
	}
	for (int i = 0; i < outR.size(); i++) {
		cout << outR[i] << ' ';
	}
}
int main(int argc, char* argv[]) {
	cin >> n; tab.resize(n); 
	for (int i = 0; i < n; i++) cin >> tab[i];
	for (int i = 0; i < n; i++) {
		if (i % 2 != 0)
			outL.pb(tab[i]);
		else
			outR.pb(tab[i]);
	}
	if (n % 2 == 0) f();
	else ff();
	
	return 0;
}