B. School Marks

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Vova studies programming in an elite school. Vova and his classmates are supposed to write n progress tests, for each test they will get a mark from 1 to p. Vova is very smart and he can write every test for any mark, but he doesn't want to stand out from the crowd too much. If the sum of his marks for all tests exceeds value x, then his classmates notice how smart he is and start distracting him asking to let them copy his homework. And if the median of his marks will be lower than y points (the definition of a median is given in the notes), then his mom will decide that he gets too many bad marks and forbid him to play computer games.

Vova has already wrote k tests and got marks a_1, …, _a__k. He doesn't want to get into the first or the second situation described above and now he needs to determine which marks he needs to get for the remaining tests. Help him do that.

Input

The first line contains 5 space-separated integers: n, k, p, x and y (1 ≤ n ≤ 999, n is odd, 0 ≤ k < n, 1 ≤ p ≤ 1000, n ≤ x ≤ n·p,1 ≤ y ≤ p). Here n is the number of tests that Vova is planned to write, k is the number of tests he has already written, p is the maximum possible mark for a test, x is the maximum total number of points so that the classmates don't yet disturb Vova, y is the minimum median point so that mom still lets him play computer games.

The second line contains k space-separated integers: a_1, …, _a__k (1 ≤ a__i ≤ p) — the marks that Vova got for the tests he has already written.

Output

If Vova cannot achieve the desired result, print "-1".

Otherwise, print n - k space-separated integers — the marks that Vova should get for the remaining tests. If there are multiple possible solutions, print any of them.

Sample test(s)

input

5 3 5 18 4
3 5 4

output

4 1

input

5 3 5 16 4
5 5 5

output

-1

Note

The median of sequence a_1, …, _a__n where n is odd (in this problem n is always odd) is the element staying on (n + 1) / 2 position in the sorted list of a__i.

In the first sample the sum of marks equals 3 + 5 + 4 + 4 + 1 = 17, what doesn't exceed 18, that means that Vova won't be disturbed by his classmates. And the median point of the sequence {1, 3, 4, 4, 5} equals to 4, that isn't less than 4, so his mom lets him play computer games.

Please note that you do not have to maximize the sum of marks or the median mark. Any of the answers: "4 2", "2 4", "5 1", "1 5", "4 1", "1 4" for the first test is correct.

In the second sample Vova got three '5' marks, so even if he gets two '1' marks, the sum of marks will be 17, that is more than the required value of 16. So, the answer to this test is "-1".

题目大意：

n是n次测试

k是已经考试了k 次

p是最高分数（这题中没有用）

x是n次总成绩的总和

y是这几次成绩的中数

求在剩下的几次测试中，vova都应该考多少分才能满足x和y.（随意的序列都可以）

分析：

根据题目中所说的  至少有(n+1)/2次要>=y

所以我们就先补y然后剩下的全部是1   这样加起来是最小的   如果他还是大与x说明不能满足

如果把剩下的都补成y还是不够(n+1)/2的话也是不能满足的。

#include
#include
#include
#include
#include
#include

using namespace std;
#define N 1050
int main()
{
int n,k,p,x,y,a[N],i;
while(scanf("%d %d %d %d %d",&n,&k,&p,&x,&y)!=EOF)
{
int Y=,sum=;
for(i=;i<=k;i++) { scanf("%d",&a[i]); if(a[i]>=y)
Y++;
sum+=a[i];
}
Y=(n+)/-Y;
if(Y<=) Y=; int X=n-Y-k; if(X<=) X=; if((sum+Y*y+X)>x || Y>(n-k))
{
printf("-1\n");
}
else
{
for(i=;i<Y;i++)
printf("%d%c",y,(x== && i==y-)?'\n':' ');
for(i=;i<X;i++)
printf("1%c",i==X-?'\n':' ');
}
}
return ;
}