codeforces 808 C. Tea Party

C. Tea Party

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp invited all his friends to the tea party to celebrate the holiday. He has n cups, one for each of his n friends, with volumes a1, a2, ..., an. His teapot stores w milliliters of tea (w ≤ a1 + a2 + ... + an). Polycarp wants to pour tea in cups in such a way that:

Every cup will contain tea for at least half of its volume
Every cup will contain integer number of milliliters of tea
All the tea from the teapot will be poured into cups
All friends will be satisfied.

Friend with cup i won't be satisfied, if there exists such cup j that cup i contains less tea than cup j but ai > aj.

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For each cup output how many milliliters of tea should be poured in it. If it's impossible to pour all the tea and satisfy all conditions then output -1.

Input

The first line contains two integer numbers n and w (1 ≤ n ≤ 100, ).

The second line contains n numbers a1, a2, ..., an (1 ≤ ai ≤ 100).

Output

Output how many milliliters of tea every cup should contain. If there are multiple answers, print any of them.

If it's impossible to pour all the tea and satisfy all conditions then output -1.

Examples

input

Copy

2 10
8 7

output

Copy

6 4

input

Copy

4 4
1 1 1 1

output

Copy

1 1 1 1

input

Copy

3 10
9 8 10

output

Copy

-1

Note

In the third example you should pour to the first cup at least 5 milliliters, to the second one at least 4, to the third one at least 5. It sums up to 14, which is greater than 10 milliliters available.

题意：

你有一个容量为w的茶壶，需要给n个朋友的水杯中倒水，i朋友的水杯的容量为ai

需要满足5个条件才个倒水成功:

1.每个人水杯中的水不小于容量的一半 2.每个人的水杯中的水是整数 3.茶壶中的水全部倒完 4.如果i容量比j容量大，但i中的水却比j中的少，i朋友会不满意需要让所有人满意

思路：

首先每个人初始的水量是容量的一半如果是奇数则 wi/2+1,然后按水杯容量排序，按容量从小到大顺序开始倒满，直到茶壶中的水倒完。

#include <cstdio>
#include <algorithm>
#include <math.h>
#include <string.h>
#include <iostream>
#define lll long long
#define INF 0x3f3f3f
using namespace std;
const int N=555;

struct node
{
    int id;
    int w;
    int ans;
}q[N];

bool cmp(node a,node b)
{
    return a.w<b.w;
}

bool cmp1(node a,node b)
{
    return a.id<b.id;
}
int main()
{
    int n,w,i,remainder;
    scanf("%d%d",&n,&w);
    for(i=0;i<n;i++)
    {
        scanf("%d",&q[i].w);
        q[i].id=i;
        if(q[i].w%2==0) q[i].ans=q[i].w/2;
        else q[i].ans=q[i].w/2+1;
        w-=q[i].ans;
    }
    sort(q,q+n,cmp);
    if(w<0) printf("-1\n");
    else
    {
        for(i=n-1;i>=0;i--)
        {
            remainder=q[i].w-q[i].ans;
            if(w>=remainder)
            {
                q[i].ans=q[i].w;
                w-=remainder;
            }
            else
            {
                q[i].ans+=w;
                w=0;
            }
            if(w==0) break;
        }
        sort(q,q+n,cmp1);
        for(i=0;i<n;i++)
            printf("%d ",q[i].ans);
    }
    return 0;
}

codeforces 808 C. Tea Party

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