Problem: 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 4
3 5 4


Output

4 1


Input

5 3 5 16 4
5 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".

Solution (in JAVA):

Solution:



import java.io.*;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.util.Collections;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.*;

public class Main1 {




    public static void main(String[] args) throws IOException {

//        try {
       // Scanner in = new Scanner(System.in) ;
         FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);


        int n = in.nextInt() ;
        int k = in.nextInt() ;
        int p = in.nextInt() ;
        int x = in.nextInt() ;
        int y = in.nextInt() ;

        ArrayList<Integer> list = new ArrayList<>() ;
        for (int i = 0; i <k ; i++) {
            list.add(in.nextInt()) ;
        }
        Collections.sort(list);
        if (k == 0){
            int mid = n/2 ;
            for (int i = 0 ; i <mid ; i++)list.add(1) ;
            for (int i = mid; i <n ; i++) {
                list.add(y) ;
            }
            int total = 0 ;
            for(int i : list)total += i ;

            if (total > x){
                System.out.println(-1);
            }
            else {
                for (int i = k; i <n ; i++) {
                    System.out.print(list.get(i) + " ");
                }
            }
            return;


        }

        int mid = n/2 ;
        if (k < mid){
            int ele = (list.get(list.size() -1 ) ) ;
            for (int i=mid-1 ; i>=k ; i--)list.add(1) ;
            if (ele >= y){
                list.add(1) ;
                mid++ ;
            }
            for(int i=mid ; i<n ; i++)list.add(y) ;

        }
        else {
            int ind = k-1 ;
            for (int i=k-1 ; i >=0 ; i--){
                if (list.get(i) < y)break;
                else ind = i ;
            }
            if (ind > mid || list.get(ind) <y){
                System.out.println(-1);
                return;
            }
            int can_add = min(n-k , mid - ind) ;
            while (can_add -- > 0)list.add(1  ) ;
            int rem = n - list.size() ;
            while (rem-- > 0)list.add(y) ;
        }

            int total = 0 ;
            for(int i : list)total += i ;

            if (total > x){
                System.out.println(-1);
            }
            else {
                for (int i = k; i <n ; i++) {
                    System.out.print(list.get(i) + " ");
                }
            }





        out.flush();
        out.close();

    }
//        catch (Exception e){
//            return;
//        }

//    }

    static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

//    static ArrayList<Integer> list = new ArrayList<>();
//    static boolean A[] = new boolean[2 * 90000001];
//
//
//    static void seive(int n) {
//        int maxn = n;
//        //int maxn = 1000000  ;
//        A[0] = A[1] = true;
//
//        for (int i = 2; i * i <= maxn; i++) {
//            if (!A[i]) {
//                for (int j = i * i; j <= maxn; j += i)
//                    A[j] = true;
//            }
//        }
//
//        for (int i = 2; i <= maxn; i++)
//            if (!A[i])
//                list.add(i);
//
//    }

    static int findLCA(int a, int b, int par[][], int depth[]) {
        if (depth[a] > depth[b]) {
            a = a ^ b;
            b = a ^ b;
            a = a ^ b;
        }
        int diff = depth[b] - depth[a];
        for (int i = 19; i >= 0; i--) {
            if ((diff & (1 << i)) > 0) {
                b = par[b][i];
            }
        }
        if (a == b)
            return a;
        for (int i = 19; i >= 0; i--) {
            if (par[b][i] != par[a][i]) {
                b = par[b][i];
                a = par[a][i];
            }
        }
        return par[a][0];
    }

    static int gcd(int a, int b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    static void formArrayForBinaryLifting(int n, int par[][]) {
        for (int j = 1; j < 20; j++) {
            for (int i = 0; i < n; i++) {
                if (par[i][j - 1] == -1)
                    continue;
                par[i][j] = par[par[i][j - 1]][j - 1];
            }
        }
    }


    static void sort(int ar[]) {
        int n = ar.length;
        ArrayList<Integer> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static void sort1(long ar[]) {
        int n = ar.length;
        ArrayList<Long> a = new ArrayList<>();
        for (int i = 0; i < n; i++)
            a.add(ar[i]);
        Collections.sort(a);
        for (int i = 0; i < n; i++)
            ar[i] = a.get(i);
    }

    static long ncr(long n, long r, long mod) {
        if (r == 0)
            return 1;
        long val = ncr(n - 1, r - 1, mod);
        val = (n * val) % mod;
        val = (val * modInverse(r, mod)) % mod;
        return val;
    }

    static long fast_pow(long base, long n, long M) {
        if (n == 0)
            return 1;
        if (n == 1)
            return base % M;
        long halfn = fast_pow(base, n / 2, M);
        if (n % 2 == 0)
            return (halfn * halfn) % M;
        else
            return (((halfn * halfn) % M) * base) % M;
    }

    static long modInverse(long n, long M) {
        return fast_pow(n, M - 2, M);
    }


    private static int countDigits(long l) {
        if (l >= 1000000000000000000L) return 19;
        if (l >= 100000000000000000L) return 18;
        if (l >= 10000000000000000L) return 17;
        if (l >= 1000000000000000L) return 16;
        if (l >= 100000000000000L) return 15;
        if (l >= 10000000000000L) return 14;
        if (l >= 1000000000000L) return 13;
        if (l >= 100000000000L) return 12;
        if (l >= 10000000000L) return 11;
        if (l >= 1000000000L) return 10;
        if (l >= 100000000L) return 9;
        if (l >= 10000000L) return 8;
        if (l >= 1000000L) return 7;
        if (l >= 100000L) return 6;
        if (l >= 10000L) return 5;
        if (l >= 1000L) return 4;
        if (l >= 100L) return 3;
        if (l >= 10L) return 2;
        return 1;
    }


    static class FastOutput implements AutoCloseable, Closeable, Appendable {
        private static final int THRESHOLD = 32 << 10;
        private final Writer os;
        private StringBuilder cache = new StringBuilder(THRESHOLD * 2);

        public FastOutput append(CharSequence csq) {
            cache.append(csq);
            return this;
        }

        public FastOutput append(CharSequence csq, int start, int end) {
            cache.append(csq, start, end);
            return this;
        }

        private void afterWrite() {
            if (cache.length() < THRESHOLD) {
                return;
            }
            flush();
        }

        public FastOutput(Writer os) {
            this.os = os;
        }

        public FastOutput(OutputStream os) {
            this(new OutputStreamWriter(os));
        }

        public FastOutput append(char c) {
            cache.append(c);
            afterWrite();
            return this;
        }

        public FastOutput append(long c) {
            cache.append(c);
            afterWrite();
            return this;
        }

        public FastOutput append(String c) {
            cache.append(c);
            afterWrite();
            return this;
        }

        public FastOutput println() {
            return append(System.lineSeparator());
        }

        public FastOutput flush() {
            try {
                //            boolean success = false;
                //            if (stringBuilderValueField != null) {
                //                try {
                //                    char[] value = (char[]) stringBuilderValueField.get(cache);
                //                    os.write(value, 0, cache.length());
                //                    success = true;
                //                } catch (Exception e) {
                //                }
                //            }
                //            if (!success) {
                os.append(cache);
                //            }
                os.flush();
                cache.setLength(0);
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
            return this;
        }

        public void close() {
            flush();
            try {
                os.close();
            } catch (IOException e) {
                throw new UncheckedIOException(e);
            }
        }

        public String toString() {
            return cache.toString();
        }

    }


    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }


        long nextLong() {
            return Long.parseLong(next());
        }
    }


}