Codeforces Round #374 (Div. 2) D - Maxim and Array

D. Maxim and Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Recently Maxim has found an array of n integers, needed by no one. He immediately come up with idea of changing it: he invented positive integerx and decided to add or subtract it from arbitrary array elements. Formally, by applying single operation Maxim chooses integeri (1 ≤ i ≤ n) and replaces thei-th element of array a_i either with a_i + x or with a_i - x. Please note that the operation may be applied more than once to the same position.

Maxim is a curious minimalis, thus he wants to know what is the minimum value that the product of all array elements (i.e.) can reach, if Maxim would apply no more thank operations to it. Please help him in that.

Input

The first line of the input contains three integers n, k andx (1 ≤ n, k ≤ 200 000, 1 ≤ x ≤ 10⁹) — the number of elements in the array, the maximum number of operations and the number invented by Maxim, respectively.

The second line contains n integers a₁, a₂, ..., a_n () — the elements of the array found by Maxim.

Output

Print n integers b₁, b₂, ..., b_n in the only line — the array elements after applying no more thank operations to the array. In particular, should stay true for every 1 ≤ i ≤ n, but the product of all array elements should beminimum possible.

If there are multiple answers, print any of them.

Examples

Input

5 3 15 4 3 5 2

Output

5 4 3 5 -1 

Input

5 3 15 4 3 5 5

Output

5 4 0 5 5 

Input

5 3 15 4 4 5 5

Output

5 1 4 5 5 

Input

3 2 75 4 2

Output

5 11 -5 

题意:

有n个数，有k次操作，每次操作能选一个数+或者-x；

要求n个数的乘积最小；

思路：

贪心；

当负数为偶数个数的时候：

　　　绝对值最小的值为正数：　－ｘ

　　　绝对值最小的值为负数：　＋ｘ

当负数为奇数个数的时候：

　　　绝对值最小的值为正数：　＋ｘ

　　　绝对值最小的值为负数：　－ｘ

代码：

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>using namespace std;#define LL __int64struct node {    int id;    LL num;    bool operator < (const node &a)const{        return abs(num)>abs(a.num);    }};priority_queue <node>Q;const int maxn =200005;LL data[maxn];int main(){    int n,k,x;    scanf("%d%d%d",&n,&k,&x);    int i,j;    int flag=1;    for(i=1;i<=n;i++){        scanf("%I64d",&data[i]);        if(data[i]<0)            flag=-flag;        Q.push( (node){i,data[i]}  );    }    for(i=1;i<=k;i++){        node nflag=Q.top();        Q.pop();        if(nflag.num<0){            if(flag<0){                nflag.num-=x;                data[nflag.id]-=x;            }            else{                nflag.num+=x;                data[nflag.id]+=x;                if(nflag.num>=0)                    flag*=-1;            }        }        else{            if(flag<0){                nflag.num+=x;                data[nflag.id]+=x;            }            else{                nflag.num-=x;                data[nflag.id]-=x;                if(nflag.num<0)                    flag*=-1;            }        }        Q.push(nflag);    }    for(i=1;i<=n;i++)        printf("%I64d ",data[i]);    printf("\n");    return 0;}