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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define N 100005

int n, m, arr[N];
int f(int);
inline int dis(int, int);

int main(){
	srand(time(NULL));
	ios_base::sync_with_stdio(0);cin.tie(0);
	cin>>n>>m;
	for(int i=0;i<n;i++) cin>>arr[i];
	double step=m, tt=0.9;
	int x=rand()%m + 1, y=f(x);
	while(step>1){
		int newX = max(0, int(x+((rand()&1)?1:-1)*rand()/(3*step))%m), newY=f(newX);
		if(newY < y){
			x = newX; 
			y = newY;
		}
		step*=tt;
	}
	cout<<y<<'\n';
	return 0;
}

inline int dis(int p, int q){
	return q>=p? q - p : q + m - p;
}

int f(int b){
	int r=0;
	for(int i=1;i<n;i++)
		r += min(dis(arr[i-1], arr[i]), dis(b, arr[i])+1);
	return r;
}