Young Timofey has a birthday today! He got kit of n cubes as a birthday PResent from his parents. Every cube has a number a_i, which is written on it. Timofey put all the cubes in a row and went to unpack other presents.

In this time, Timofey's elder brother, Dima reordered the cubes using the following rule. Suppose the cubes are numbered from 1 to n in their order. Dima performs several steps, on step i he reverses the segment of cubes from i-th to (n?-?i?+?1)-th. He does this while i?≤?n?-?i?+?1.

After performing the Operations Dima went away, being very proud of himself. When Timofey returned to his cubes, he understood that their order was changed. Help Timofey as fast as you can and save the holiday — restore the initial order of the cubes using information of their current location.

The first line contains single integer n (1?≤?n?≤?2·10⁵) — the number of cubes.

The second line contains n integers a₁,?a₂,?...,?a_n (?-?10⁹?≤?a_i?≤?10⁹), where a_i is the number written on the i-th cube after Dima has changed their order.

Print n integers, separated by spaces — the numbers written on the cubes in their initial order.

It can be shown that the answer is unique.

74 3 7 6 9 1 2Output
2 3 9 6 7 1 4Input
86 1 4 2 5 6 9 2Output
2 1 6 2 5 4 9 6Note
Consider the first sample.
At the begining row was [2, 3, 9, 6, 7, 1, 4].After first operation row was [4, 1, 7, 6, 9, 3, 2].After second operation row was [4, 3, 9, 6, 7, 1, 2].After third operation row was [4, 3, 7, 6, 9, 1, 2].
At fourth operation we reverse just middle element, so nothing has changed. The final row is [4, 3, 7, 6, 9, 1, 2]. So the answer for this case is row [2, 3, 9, 6, 7, 1, 4].題目大意：對(duì)于給定的一個(gè)序列進(jìn)行操作，序列長(zhǎng)度為n，進(jìn)行操作，第i步操作即是把從第i起到第n-i+1的數(shù)進(jìn)行翻轉(zhuǎn)，保證2*i<=n+1;并且進(jìn)行輸出題目分析：由于答案不唯一，進(jìn)行模擬運(yùn)算可以發(fā)現(xiàn)，第i個(gè)數(shù)若為偶數(shù)，即和第n-i+1的數(shù)進(jìn)行調(diào)換，否則就保持不變。
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 100005*2;long long a[maxn],b[maxn];int main(){	int n;	long long k;	while((scanf("%d",&n))!=EOF){		int i;		for(i=1;i<=n;i++)		   scanf("%lld",&a[i]);		   i=1;        while(i<=n-i+1){        	if(i%2 == 1)         swap(a[i],a[n-i+1]);         i++;		}		for(i=1;i<n;i++){			printf("%lld ",a[i]);		}		printf("%lld/n",a[n]);	}	return 0;}