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.

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



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

k space-separated integers: a1, ..., ak (1 ≤ ai ≤ p) — the marks that Vova got for the tests he has already written.



Output



-1".

n - k



Examples



input



5 3 5 18 4 3 5 4



output



4 1



input



5 3 5 16 4 5 5 5



output



-1






本题的题意是给出n个测试k个成绩，补全n-k个成绩

要求是所有成绩之和不能大于x 并且成绩排序之后的中位数不能小于 y

思路是将数据分成两半 一半是大于y的并且数量是n/2+1，另一半是小于y的

只需要按照数量来构造就行

代码如下：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
#define N   1005
int a[N],b[N];
int main()
{
  int i,j,n,k,x,y,p,pp,sum;
  while(scanf("%d %d %d %d %d",&n,&k,&p,&x,&y)!=EOF)
  {
    sum=0;
    memset(a,0,sizeof(a));
    memset(b,0,sizeof(b));
    int l=0,r=0;
    for(i=1;i<=k;i++)
    {
    cin>>a[i];
    sum+=a[i];
    if(a[i]<y) l++;
    else r++;
    }
    if(sum>x||sum+n-k>x)
    {
      cout<<"-1"<<endl;
      continue;
    }
    int half=n/2+1;
    if(l>=half)
    {
      cout<<"-1"<<endl;
      continue;
    } 
  i=1;
    if(r<=half)
    {
       pp=half-r;
      for(i=1;i<=pp;i++)
      {
      b[i]=y;
      sum+=b[i];  
        }
    }
    int jj=n-l-r-i+1;
        while(jj)
    {
      b[i]=1;
      sum++;
      jj--;
      i++;
    }
     if(sum>x)
      cout<<"-1"<<endl;
     
       else {
        for(i=1;i<=n-k;i++)
        cout<<b[i]<<" ";
        cout<<endl;
     }
  }
  return 0;
}