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 <algorithm>
#include <cassert>
#include <climits>
#include <cmath>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <vector>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef pair<int, int> ipair;
typedef tuple<int, int, int> ituple;

// const ll INF = LLONG_MAX;
// const int MOD = (int)1e9 + 7;
// const double EPS = 1e-10;
#define PI acosl(-1)
#define MAX_N 200000 + 2

void exec(){
  int n;
  int a[MAX_N], b[MAX_N];

  cin >> n;
  for (int i = 0; i < n; i++){
    scanf("%d", &a[i]);
  }

  int start = 0;
  int end = n - 1;
  bool isStart = true;
  int currentIndex = n - 1;

  while (currentIndex >= 0){
    if (isStart){
      b[start] = a[currentIndex];
      // printf("start: b[%d] = %d\n", start, a[currentIndex]);
      start++;
    }
    else{
      b[end] = a[currentIndex];
      // printf("end: b[%d] = %d\n", end, a[currentIndex]);
      end--;
    }

    isStart = !isStart;
    currentIndex -= 1;
  }

  for (int i = 0; i < n; i++){
    printf("%d%c", b[i], i == n-1? '\n' : ' ');
  }

}

void solve(){
  int t = 1;
  // scanf("%d", &t);
  for (int i = 0; i < t; i++){
    exec();
  }
}

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