Nick had received an awesome array of integers a=[a1,a2,…,an] as a gift for his 5 birthday from his mother. He was already going to explore its various properties but after unpacking he was disappointed a lot because the product a1⋅a2⋅…an of its elements seemed to him not large enough.
He was ready to throw out the array, but his mother reassured him. She told him, that array would not be spoiled after the following operation: choose any index i (1≤i≤n) and do ai:=−ai−1.
For example, he can change array [3,−1,−4,1] to an array [−4,−1,3,1] after applying this operation to elements with indices i=1 and i=3.
Kolya had immediately understood that sometimes it’s possible to increase the product of integers of the array a lot. Now he has decided that he wants to get an array with the maximal possible product of integers using only this operation with its elements (possibly zero, one or more times, as many as he wants), it is not forbidden to do this operation several times for the same index.
Help Kolya and print the array with the maximal possible product of elements a1⋅a2⋅…an which can be received using only this operation in some order.
If there are multiple answers, print any of them.
Input
The first line contains integer n (1≤n≤105) — number of integers in the array.
The second line contains n integers a1,a2,…,an (−106≤ai≤106) — elements of the array
Output
Print n numbers — elements of the array with the maximal possible product of elements which can be received using only this operation in some order from the given array.
If there are multiple answers, print any of them.
Examples
Input
4
2 2 2 2
Output
-3 -3 -3 -3
Input
1
0
Output
0
Input
3
-3 -3 2
Output
-3 -3 2
题意：
给出一个数组，然后执行操作。。。。(省略62个字）任性？！？！？！
思路：
先让所有的数都按照操作变成负数（因为改变过的负数的绝对值要比正数的绝对值大）；然后判断n的奇偶性，如果n为偶数，那么n个负数的乘积一定为正数，直接输出就行，如果n为奇数，那么把绝对值最小的那个负数变成整数就行了；
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
int a[100010];
int main()
{
int n;
cin >>n;
int minn=INF;
for(int i=0;i<n;i++)
{
cin >>a[i];
if(a[i]>=0) a[i]=-1*a[i]-1;
minn=min(minn,a[i]);
}
int flag=1;
for(int i=0;i<n;i++)
{
if(a[i]==minn&&n%2==1&&flag) cout <<-1*a[i]-1<<" ",flag=0;
else cout <<a[i]<<" ";
}
}
