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 <stdio.h>
#include <algorithm>
#include <iostream>
#include <vector>
#include <assert.h>
#include <map>
#include <set>
#include <unordered_set>
#include <unordered_map>
#include <stdlib.h>
#include <math.h>
#include <string>
#include <time.h>
#include <string.h>
#include <queue>
#include <complex>
#include <iomanip>
#include <stack>
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
ll mod=1e9+7;
double eps=1e-7;
ll exp(ll x,ll y){if(y<0) return 0; ll ret=1;for(;y;y>>=1,x=(x*x)%mod){if(y&1)ret=(ret*x)%mod;}return ret;}
ull pexp(ull x,ull y){if(y<0) return 0; ull ret=1; for(;y;y>>=1,x=(x*x)){if(y&1)ret=(ret*x);}return ret;}
ll gcd(ll x,ll y){if(!x||!y) return x+y; return x%y==0?y:gcd(y,x%y);}
ll lcm(ll x,ll y){return x*(y/gcd(x,y));}
ll bsum(ll u,ll b){ll ret=0;if(u<b)return u;while(u){ret+=u%b;u/=b;}return ret;}
ll prival(ll u,ll p){ll cn=0;while(u%p==0){cn++;u=u/p;}return cn;}
ll minv(ll a,ll b){return 1<a?b-minv(b%a,a)*b/a:1;}
ll extm(ll a,ll b){ll ret=0;while(a!=0){if(a%2==1){ret+=b;ret%=mod;}a>>=1;b=(2*b)%mod;}return ret;}   
ll eaphi(ll x){ll t=x,ret=x,i;for(i=2;i*i<=x;i++){if(t%i==0){ret-=ret/i;while(t%i==0) t/=i;}}if(t!=1) ret-=ret/t;return ret;}
ll eadivc(ll x){ll ret=0;ll i;for(i=1;i*i<=x;i++){if(x%i==0 && i*i!=x) ret+=2;if(x%i==0 && i*i==x) ret+=1;}return ret;}
ll eadivs(ll x){ll ret=0;ll i;for(i=1;i*i<=x;i++){if(x%i==0 && i*i!=x) ret+=i+x/i;if(x%i==0 && i*i==x) ret+=i;}return ret;}
ll ndig(ll x, ll b){ll ret=0;while(x){x/=b; ret++;}return ret;}
ll rev(ll n, ll b){ll ret=0;while(n){ret=b*ret+n%b; n/=b;}return ret;}
ll sq(ll x){ll t=(ll)sqrt(x); for(ll i=t-2 ; i<=t+2 ; i++) if(i*i==x) return abs(i); return -1;}
ll extexp(ll x,ll y){if(y<0) return 0; ll ret=1;for(;y;y>>=1,x=extm(x,x)){if(y&1)ret=extm(ret,x);}return ret;}
bool isprime(ll x){if(x<=1) return false; for(ll i=2;i*i<=x;i++){if(x%i==0){return false;}}return true;}
deque<int> Q;
int n;

int main(void)
{
	ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	int i, x; cin>>n;
	for(i=1 ; i<=n ; i++)
	{
		cin>>x;
		if(i%2==1) Q.push_front(x);
		if(i%2==0) Q.push_back(x);
	}
	if(n%2==1)
	{
		for(i=1 ; i<=n ; i++)
		{
			cout<<Q.front()<<" ";
			Q.pop_front();
		}
	}
	if(n%2==0)
	{
		for(i=1 ; i<=n ; i++)
		{
			cout<<Q.back()<<" ";
			Q.pop_back();
		}
	}
	return 0;
}