Danang and Darto are classmates. They are given homework to create a permutation of N integers from 1 to N. Danang has completed the homework and created a permutation A of N integers. Darto wants to copy Danang's homework, but Danang asks Darto to change it up a bit so it does not look obvious that Darto copied.

The difference of two permutations of N integers A and B, denoted by diff(A,B), is the sum of the absolute difference of Ai and Bi for all i. In other words, diff(A,B)=ΣNi=1|Ai−Bi|. Darto would like to create a permutation of N integers that maximizes its difference with A. Formally, he wants to find a permutation of N integers Bmax such that diff(A,Bmax)≥diff(A,B′) for all permutation of N integers B′.

Darto needs your help! Since the teacher giving the homework is lenient, any permutation of N integers B is considered different with A if the difference of A and B is at least N. Therefore, you are allowed to return any permutation of N integers B such that diff(A,B)≥N.

Of course, you can still return Bmax if you want, since it can be proven that diff(A,Bmax)≥N for any permutation A and N>1. This also proves that there exists a solution for any permutation of N integers A. If there is more than one valid solution, you can output any of them.

Input

Input begins with a line containing an integer: N (2≤N≤100000) representing the size of Danang's permutation. The next line contains N integers: Ai (1≤Ai≤N) representing Danang's permutation. It is guaranteed that all elements in A are distinct.

Output

Output in a line N integers (each separated by a single space) representing the permutation of N integers B such that diff(A,B)≥N. As a reminder, all elements in the permutation must be between 1 to N and distinct.

Examples

input

4
1 3 2 4
output

4 2 3 1
input

2
2 1
output

1 2
Note

Explanation for the sample input/output #1

With A=[1,3,2,4] and B=[4,2,3,1], diff(A,B)=|1−4|+|3−2|+|2−3|+|4−1|=3+1+1+3=8. Since 8≥4, [4,2,3,1] is one of the valid output for this sample.

解题思路j:题目要求的是将数组元素重新排列,使得对应位置上的元素相减和的绝对值最大,即保证最大程度每一位上的元素都不同,将第一位对应最后一位,第二位对应倒数第二位,...。

AC代码:

#include <iostream>
using namespace std;
int main()
{
int n,a;
while(cin>>n)
{
for(int i=;i<=n;i++)
{
cin>>a;
cout<<n-a+<<" ";
}
cout<<endl;
}
return ;
}

2019-2020 ICPC, Asia Jakarta Regional Contest A. Copying Homework的更多相关文章

  1. 2019-2020 ICPC, Asia Jakarta Regional Contest A. Copying Homework (思维)

    Danang and Darto are classmates. They are given homework to create a permutation of N integers from  ...

  2. 2019-2020 ICPC, Asia Jakarta Regional Contest (Online Mirror, ICPC Rules, Teams Preferred)

    2019-2020 ICPC, Asia Jakarta Regional Contest (Online Mirror, ICPC Rules, Teams Preferred) easy: ACE ...

  3. 2019-2020 ICPC, Asia Jakarta Regional Contest

    目录 Contest Info Solutions A. Copying Homework C. Even Path E. Songwriter G. Performance Review H. Tw ...

  4. 2019-2020 ICPC, Asia Jakarta Regional Contest H. Twin Buildings

    As you might already know, space has always been a problem in ICPC Jakarta. To cope with this, ICPC ...

  5. 2018 ICPC Asia Jakarta Regional Contest

    题目传送门 题号 A B C D E F G H I J K L 状态 Ο . . Ο . . Ø Ø Ø Ø . Ο Ο:当场 Ø:已补 .  :  待补 A. Edit Distance Thin ...

  6. 2019-2020 ICPC, Asia Jakarta Regional Contest C. Even Path

    Pathfinding is a task of finding a route between two points. It often appears in many problems. For ...

  7. 模拟赛小结:2019-2020 ICPC, Asia Jakarta Regional Contest

    比赛链接:传送门 离金最近的一次?,lh大佬carry场. Problem A. Copying Homework 00:17(+) Solved by Dancepted 签到,读题有点慢了.而且配 ...

  8. 2019-2020 ICPC, Asia Jakarta Regional Contest C. Even Path(思维)

    Pathfinding is a task of finding a route between two points. It often appears in many problems. For ...

  9. Asia Jakarta Regional Contest 2019 I - Mission Possible

    cf的地址 因为校强, "咕咕十段"队获得了EC-final的参赛资格 因为我弱, "咕咕十段"队现在银面很大 于是咕咕十段决定进行训练. 周末vp了一场, 这 ...

随机推荐

  1. sql防止注入的技巧

    from Stack Overflow Here is a similar solution which I think is more efficient in building up the li ...

  2. Codeforces Round #512 (Div. 2, based on Technocup 2019 Elimination Round 1) E. Vasya and Good Sequences(DP)

    题目链接:http://codeforces.com/contest/1058/problem/E 题意:给出 n 个数,对于一个选定的区间,区间内的数可以通过重新排列二进制数的位置得到一个新的数,问 ...

  3. Java日期工具类DateUtils详解(转)

    jar包 appache下的 common-lang3 一. 对指定的日期新增年.月.周.日.小时.分钟.秒.毫秒 public static Date addDays(Date date, int ...

  4. 关于dll注入

    例如:     有一个游戏修改器:其中有一个按钮“自动打怪”:点击时游戏会实现相应的功能:     对于游戏程序来说,自动打怪操作本质上就是call调用一个函数:     但是修改器和游戏是两个独立的 ...

  5. use potato

  6. 想学习linux操作系统,于是选择了在win8 虚拟机VM player 里装了Linux版本Centos7

    第一次接触linux,第一次玩linux的命令行哈. 以下python使用的都是自带的python2.x版本 先新建一个简单的文件夹py_studydir 新建该文件夹下面的一个py文件 写入pyth ...

  7. 树莓派安装alsa-lib库

    安装alsa-lib库 apt-get install libasound2-dev dpkg -L libasound2-dev 参考:https://blog.csdn.net/happygril ...

  8. vue 源码解析computed

    计算属性 VS 侦听属性 Vue 的组件对象支持了计算属性 computed 和侦听属性 watch 2 个选项,很多同学不了解什么时候该用 computed 什么时候该用 watch.先不回答这个问 ...

  9. UVA 796 Critical Links —— (求割边(桥))

    和求割点类似,只要把>=改成>即可.这里想解释一下的是,无向图没有重边,怎么可以使得low[v]=dfn[u]呢?只要它们之间再来一个点即可. 总感觉图论要很仔细地想啊- -一不小心就弄混 ...

  10. HDFS CheckPoint && SavePoint

    HDFS CheckPoint && SavePoint 标签(空格分隔): Hadoop HDFS CheckPoint HDFS 将文件系统的元数据信息存放在 fsimage 和一 ...