【Codeforces 808C】【贪心】Tea Party题解

C. Tea Party

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

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

2 10

8 7

output

6 4

input

4 4

1 1 1 1

output

1 1 1 1

input

3 10

9 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.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <stack>
#define INF 2100000000
#define LL long long
#define clr(x) memset(x, 0, sizeof(x))
#define ms(a, x) memset(x, a, sizeof(x))

using namespace std;

template <class T> inline void read(T &x) {
int flag = 1; x = 0;
char ch = (char)getchar();
while(ch <  '0' || ch >  '9') { if(ch == '-')  flag = -1; ch = (char)getchar(); }
while(ch >= '0' && ch <= '9') { x = (x<<1)+(x<<3)+ch-'0'; ch = (char)getchar(); }
x *= flag;
}

const int maxn = 110;
int n,w;
struct cup { int num,m,t; } a[maxn];

inline int cmp(cup x, cup y) { return x.m < y.m; }

inline int cmp2(cup x, cup y) { return x.num < y.num; }

int main() {
read(n); read(w);
for(int i = 1; i <= n; i++) read(a[i].m), a[i].num = i;
int ok = 1;
sort(a+1, a+n+1, cmp);
for(int i = 1; i <= n; i++) {
int x = (a[i].m+1)>>1;
if(w < x) { ok = 0; break; }
if(i == n) {
x = w;
if(a[i].m < x) {
w -= a[i].m-a[i].t;
a[i].t = a[i].m;
for(int j = n-1; j >= 1; j--) {
int y = a[j].m;
y = min(w+a[j].t, y);
w -= y-a[j].t;
a[j].t = y;
if(w <= 0) break;
}
if(w > 0) { ok = 0; break; }
} else a[i].t=x;
}
if(i != n) a[i].t = x;
w -= x;
}
if(!ok) { puts("-1"); return 0; }
sort(a+1, a+n+1, cmp2);
for(int i = 1; i <= n; i++) printf("%d ",a[i].t);
return 0;
}