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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>
#define REP(i, n) for(i = 0; i < n; ++i)
#define FOR(i, a, b) for(i = a; i < b; ++i)
#define BORDER 400

void yes_or_no(char boolean){
    if(boolean){
        printf("YES\n");
    }else{
        printf("NO\n");
    }
}

int main(){
    int n, i;
    scanf("%d", &n);
    int a[n];

    REP(i, n){
        scanf("%d", &a[i]);
    }

    if(n % 2){
        for(i = n - 1; i >= 0; i -= 2){
            printf("%d ", a[i]);
        }
        for(i = 1; i < n - 2; i += 2){
            printf("%d ", a[i]);
        }
        if(n > 1) printf("%d\n", a[n - 2]);

    }else{
        for(i = n - 1; i >= 0; i -= 2){
            printf("%d ", a[i]);
        }
        for(i = 0; i < n - 2; i += 2){
            printf("%d ", a[i]);
        }
        if(n > 1) printf("%d\n", a[n - 2]);
    }
    return 0;
}