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 <bits/stdc++.h>

#define INF 1 << 29
#define REP(i,n) for(int i=0; i<(int)(n); i++)
#define FOR(i,k,n) for(int i=(k);i<(int)(n);i++)

using namespace std;

typedef pair<int,int> pii;
typedef long long int ll;

void gen_sosuu(int sosuu[]){
  REP(i,100010){
    sosuu[i] = true;
  }
  sosuu[0] = false;
  sosuu[1] = false;
  sosuu[2] = true;
  FOR(i,2,sqrt(100000)+1){
    if(sosuu[i] == false) continue;
    for(int j = 2; i*j<=100000;j++){
      //printf("%d %d %d\n",i*j,i,j);
      sosuu[i*j] =false;
    }
  }
}

/*
0 6 7 6 7 0
0
6 0
7 0 6
6 6 0 7
7 7 0 6 6
0 6 6 0 7 7
*/

int n;
int a;
vector<int> vec;

int main(){
  scanf("%d",&n);
  REP(i,n){
    scanf("%d",&a);
    if(i%2==1){
      vec.insert(vec.begin(),a);
    }else{
      vec.push_back(a);
    }
  }
  if(n%2==1){
    reverse(vec.begin(),vec.end());
  }
  REP(i,n){
    printf("%d ",vec[i]);
  }
  printf("\n");
}