/*

     贪心：首先要注意，y是中位数的要求；先把其他的都设置为1，那么最多有(n-1)/2个比y小的，cnt记录比y小的个数

             num1是输出的1的个数，numy是除此之外的数都为y，此时的numy是最少需要的，这样才可能中位数大于等于y

 */

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 using namespace std;

 const int MAXN = 1e3 + ;

 const int INF = 0x3f3f3f3f;

 int a[MAXN];

 int main(void)        //Codeforces Round #301 (Div. 2) B. School Marks

 {

     //freopen ("B.in", "r", stdin);

     int n, k, p, x, y;

     while (scanf ("%d%d%d%d%d", &n, &k, &p, &x, &y) == )

     {

         int sum = , cnt = ;

         for (int i=; i<=k; ++i)

         {

             scanf ("%d", &a[i]);    sum += a[i];

             if (a[i] < y)    cnt++;

         }

         if (cnt <= n / )

         {

             int num1 = min (n /  - cnt, n - k);

             int numy = n - k - num1;

             sum += num1 + numy * y;

             if (sum > x)    puts ("-1");

             else

             {

                 for (int i=; i<=num1; ++i)    printf ("%d%c", , (numy== && i==num1) ? '\n' : ' ');

                 for (int i=; i<=numy; ++i)    printf ("%d%c", y, (i==numy) ? '\n' : ' ');

             }

         }

         else    puts ("-1");

     }

     return ;

 }