【CodeForces

E - Tea Party

 

Polycarp invited all his friends to the tea party to celebrate the holiday. He has ncups, one for each of his n friends, with volumes a₁, a₂, ..., a_n. His teapot stores wmilliliters of tea (w ≤ a₁ + a₂ + ... + a_n). 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 a_i > a_j.

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, ..., a_n (1 ≤ a_i ≤ 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.

Example

Input

2 108 7

Output

6 4 

Input

4 41 1 1 1

Output

1 1 1 1 

Input

3 109 8 10

Output

-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：给你n个容量为a1, a2...an的茶杯，现在茶壶中有w体积的茶水，要求倒完茶水后要使所有客人开心，开心的条件是所有茶杯中的茶水不小于茶杯容积的一半，且容积大的茶杯中茶水不能少于容积小的茶杯中的茶水。

分析：先判断所有茶杯中倒入一半茶水时所需的茶水是否大于茶壶中的茶水。如果大于则输出-1。先给每个茶杯中加入一半的茶水，根据题意，我么可以将茶杯的容积排序，然后将剩余的茶水倒入大容积的茶杯中。最后按照茶杯序列排序输出即可。

代码如下：

#include <map>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <string>#include <cstring>#include <iostream>#include <algorithm>#define LL long longusing namespace std;const int MX = 1e5 + 5;const int mod = 1e9 + 7;const int INF = 2e9 + 5;struct node{    int x, num, ans;}a[MX];bool cmp1(node a, node b){    return a.x < b.x;}bool cmp2(node a, node b){    return a.num < b.num;}int main() {    int n, k;    scanf("%d%d", &n, &k);    int need = 0;    for(int i = 1; i <= n; i++){        scanf("%d", &a[i].x);        a[i].num = i;        need += (a[i].x+1)/2;    }    if(need > k){        printf("-1\n");        return 0;    }    sort(a+1, a+n+1, cmp1);    for(int i = 1; i <= n; i++){        a[i].ans = (a[i].x+1)/2;        k -= a[i].ans;    }    //cout << k << endl;    for(int i = n; i >= 1; i--){        if(k >= (a[i].x - a[i].ans)){            k -= (a[i].x - a[i].ans);            a[i].ans = a[i].x;        }        else{            a[i].ans += k;            break;        }    }    sort(a+1, a+1+n, cmp2);    printf("%d", a[1].ans);    for(int i = 2; i <= n; i++){        printf(" %d", a[i].ans);    }    printf("\n");    return 0;}