CodeForces

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 a1, ..., ak. 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: a1, ..., ak (1 ≤ ai ≤ 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.

Examples

input

5 3 5 18 43 5 4

output

4 1

input

5 3 5 16 45 5 5

output

-1

Note

The median of sequence a1, ..., an 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 ai.

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".

题意： 假设小明有n场考试，已经考了k场，，剩下的几场考试  考试分数不能超过p分，，n场考试的分数不能超过x分   ，n场考试的中位数必须大于等于y分，，（n这里为奇数）

思路： 第一我们只需要找到已经考了的k场考试中是否小于y的个数已经>n/2，或者总分数是否已经超过x，，如果两者成立其一那么输出-1.

否则, 贪心，我们想要最后的分数越小越好，所以每次考试只考1分知道考试中小于y的个数==n/2  那么之后的考试都考y分，就可以保证中位数大于等于y  并且  总分数最小。

代码： 

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 10005using namespace std;int a[N];int n,k,p,s,x,y;int main(){int i,j;scanf("%d %d %d %d %d",&n,&k,&p,&x,&y);int sum,cnt;sum=cnt=0;for(i=1;i<=k;i++){scanf("%d",&a[i]);if(a[i]<y) sum++;cnt+=a[i];}if(cnt>x||sum>n/2){printf("-1\n");return 0;}if(sum==n/2){for(i=k+1;i<=n;i++){a[i]=y;cnt+=y;}if(cnt>x){printf("-1\n");}else{for(i=k+1;i<=n;i++){if(i==n) printf("%d\n",a[i]);else printf("%d ",a[i]);}}}else{for(i=k+1;i<=n;i++){if(sum==n/2) break;a[i]=1;cnt+=1;sum++;}for(;i<=n;i++){a[i]=y;cnt+=y;}if(cnt>x){printf("-1\n");}else{for(i=k+1;i<=n;i++){if(i==n) printf("%d\n",a[i]);else printf("%d ",a[i]);}}}return 0;}