Maxim and Array

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 integer x and decided to add or subtract it from arbitrary array elements. Formally, by applying single operation Maxim chooses integer i (1 ≤ i ≤ n) and replaces the i-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 than k operations to it. Please help him in that.

Input

The first line of the input contains three integers n, k and x (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 than k operations to the array. In particular,  should stay true for every 1 ≤ i ≤ n, but the product of all array elements should be minimum 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：一个序列长为n，有k个x让你去给序列任意一个数加||减x，然后要求序列各项乘积最大，输出变化后的序列

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+9;
struct num {
    ll abs,val,idx;
    num(ll val,ll abs,ll idx):val(val),abs(abs),idx(idx) {}
    bool operator < (const num& rhs)const {
        return abs>rhs.abs;
    }
};
ll a[N],n,k,x,zero,neg;
int main() {
    //freopen("f.txt","r",stdin);
    priority_queue<num>q;
    scanf("%I64d%I64d%I64d",&n,&k,&x);
    zero=neg=0;
    for(int i=1; i<=n; i++) {
        scanf("%I64d",&a[i]);
        if(a[i]==0)zero++;
        if(a[i]<0)neg++;
        q.push(num(a[i],abs(a[i]),i));
    }
    if(zero>k) {
        for(int i=1; i<=n; i++)printf("%I64d ",a[i]);
        return 0;
    }

    if(neg%2==0) {
        if(zero) {
            neg++;
            zero--,k--;
            num t=q.top();
            q.pop();
            a[t.idx]-=x;
            t.val-=x;
            t.abs+=x;
            q.push(t);
        }
    }
    while(zero--) {
        k--;
        num t=q.top();
        q.pop();
        a[t.idx]+=x;
        t.val+=x;
        t.abs+=x;
        q.push(t);
    }
    if(neg%2==0) {
        num t=q.top();
        q.pop();
        if(t.val<0) {
            while(t.val<=0&&k)t.val+=x,k--;

        } else
            while(t.val>=0&&k)t.val-=x,k--;
        a[t.idx]=t.val;
        t.abs=abs(t.val);
        q.push(t);
    }
    while(k--) {
        num t=q.top();
        q.pop();
        if(t.val<0)a[t.idx]-=x;
        else a[t.idx]+=x;
        t.abs+=x;
        q.push(t);
    }
    for(int i=1; i<=n; i++)printf("%I64d ",a[i]);
    return 0;
}

原文地址：https://www.cnblogs.com/geraldg/p/11247812.html