Problem: Olya likes milk very much. She drinks k cartons of milk each day if she has at least k and drinks all of them if she doesn't. But there's an issue — expiration dates. Each carton has a date after which you can't drink it (you still can drink it exactly at the date written on the carton). Due to this, if Olya's fridge contains a carton past its expiry date, she throws it away.

Olya hates throwing out cartons, so when she drinks a carton, she chooses the one which expires the fastest. It's easy to understand that this strategy minimizes the amount of cartons thrown out and lets her avoid it if it's even possible.

<image> Milk. Best before: 20.02.2017.

The main issue Olya has is the one of buying new cartons. Currently, there are n cartons of milk in Olya's fridge, for each one an expiration date is known (how soon does it expire, measured in days). In the shop that Olya visited there are m cartons, and the expiration date is known for each of those cartons as well.

Find the maximum number of cartons Olya can buy so that she wouldn't have to throw away any cartons. Assume that Olya drank no cartons today. 

Input

In the first line there are three integers n, m, k (1 ≤ n, m ≤ 106, 1 ≤ k ≤ n + m) — the amount of cartons in Olya's fridge, the amount of cartons in the shop and the number of cartons Olya drinks each day.

In the second line there are n integers f1, f2, ..., fn (0 ≤ fi ≤ 107) — expiration dates of the cartons in Olya's fridge. The expiration date is expressed by the number of days the drinking of this carton can be delayed. For example, a 0 expiration date means it must be drunk today, 1 — no later than tomorrow, etc.

In the third line there are m integers s1, s2, ..., sm (0 ≤ si ≤ 107) — expiration dates of the cartons in the shop in a similar format.

Output

If there's no way for Olya to drink the cartons she already has in her fridge, print -1.

Otherwise, in the first line print the maximum number x of cartons which Olya can buy so that she wouldn't have to throw a carton away. The next line should contain exactly x integers — the numbers of the cartons that should be bought (cartons are numbered in an order in which they are written in the input, starting with 1). Numbers should not repeat, but can be in arbitrary order. If there are multiple correct answers, print any of them.

Examples

Input

3 6 2
1 0 1
2 0 2 0 0 2


Output

3
1 2 3

Input

3 1 2
0 0 0
1


Output

-1

Input

2 1 2
0 1
0


Output

1
1 

Note

In the first example k = 2 and Olya has three cartons with expiry dates 0, 1 and 1 (they expire today, tomorrow and tomorrow), and the shop has 3 cartons with expiry date 0 and 3 cartons with expiry date 2. Olya can buy three cartons, for example, one with the expiry date 0 and two with expiry date 2.

In the second example all three cartons Olya owns expire today and it means she would have to throw packets away regardless of whether she buys an extra one or not.

In the third example Olya would drink k = 2 cartons today (one she alreay has in her fridge and one from the shop) and the remaining one tomorrow.

Solution (in JAVA):

Solution:

import java.io.*;
import java.util.*;

/**
 * Created by brkuma on 3/2/2017.
 */

public class CF767D {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        CF767D solver = new CF767D();
        solver.solve(in, out);
        out.flush();
        out.close();
    }

    public void solve(InputReader in, PrintWriter out) {
        // Main Code goes here
        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();
        int mx = 0;
        int[] f = in.nextInt1D(n);
        Arrays.sort(f);
        int[] s = in.nextInt1D(m);
        Pair<Integer, Integer>[] shop = new Pair[m];
        for (int i = 0; i < m; i++) {
            shop[i] = Pair.create(s[i], i + 1);
            mx = Math.max(mx, s[i]);
        }
        Arrays.sort(shop, new Comparator<Pair<Integer, Integer>>() {
            @Override
            public int compare(Pair<Integer, Integer> one, Pair<Integer, Integer> two) {
                return one.first - two.first;
            }
        });
//        if (!canFinish(f, k)) {
//            out.println(-1);
//            return;
//        }

        List<Integer> ans = new ArrayList<>();
        int ff = 0;
        int ss = 0;
        int day = 0;

        while (true) {
            int possible = k;
            // try to drink from fridge
            while (ff < n && f[ff] <= day && possible > 0) {
                possible--;
                ff++;
            }
            // check if she need to throw away some from fridge
            if (ff < n && f[ff] <= day) {
                out.println(-1);

                return;
            }
            // try to pick from shop
            while (ss < m && possible > 0) {
                if (shop[ss].first >= day) {
                    possible--;
                    ans.add(shop[ss].second);
                }
                ss++;
            }
            if (possible == k) {
                break;
            }
            day++;
        }

        Collections.sort(ans);
        out.println(ans.size());
        StringBuilder sb = new StringBuilder();
        for (int ii = 0; ii < ans.size(); ii++) {
            if (ii != 0) {
                sb.append(" ");
            }
            sb.append(ans.get(ii));
        }
        out.println(sb);
        
    }

    private boolean canFinish(int[] f, int k) {

        int n = f.length;
        int bound = n / k;
        if (n % k != 0) {
            bound++;
        }
        boolean can = true;
        for (int i = 0; i < bound; i++) {
            for (int j = 0; j < k; j++) {
                int idx = i * k + j;
                if (idx < n) {
                    if (f[idx] - i < 0) {
                        can = false;
                    }
                }
            }
        }
        return can;
    }

    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer stt;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        public String nextString() {
            while (stt == null || !stt.hasMoreTokens()) {
                stt = new StringTokenizer(nextLine());
            }
            return stt.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextString());
        }

        public long nextLong() {
            return Long.parseLong(nextString());
        }

        public double nextDouble() {
            return Double.parseDouble(nextString());
        }

        public int[] nextInt1D(int n) {
            int[] ans = new int[n];
            for (int i = 0; i < n; i++) {
                ans[i] = nextInt();
            }
            return ans;
        }
    }


    static class ArrayUtils {
        public static void fill(int[][] f, int value) {
            for (int i = 0; i < f.length; ++i) {
                Arrays.fill(f[i], value);
            }
        }
    }

    /**
     * Container to ease passing around a tuple of two objects. This object provides a sensible
     * implementation of equals(), returning true if equals() is true on each of the contained
     * objects.
     */
    static class Pair<F, S> {
        public final F first;
        public final S second;

        /**
         * Constructor for a Pair.
         *
         * @param first  the first object in the Pair
         * @param second the second object in the pair
         */
        public Pair(F first, S second) {
            this.first = first;
            this.second = second;
        }

        /**
         * Checks the two objects for equality by delegating to their respective
         * {@link Object#equals(Object)} methods.
         *
         * @param o the {@link Pair} to which this one is to be checked for equality
         * @return true if the underlying objects of the Pair are both considered
         * equal
         */
        @Override
        public boolean equals(Object o) {
            if (!(o instanceof Pair)) {
                return false;
            }
            Pair<?, ?> p = (Pair<?, ?>) o;
            return Objects.equals(p.first, first) && Objects.equals(p.second, second);
        }

        /**
         * Compute a hash code using the hash codes of the underlying objects
         *
         * @return a hashcode of the Pair
         */
        @Override
        public int hashCode() {
            return (first == null ? 0 : first.hashCode()) ^ (second == null ? 0 : second.hashCode());
        }

        /**
         * Convenience method for creating an appropriately typed pair.
         *
         * @param a the first object in the Pair
         * @param b the second object in the pair
         * @return a Pair that is templatized with the types of a and b
         */
        public static <A, B> Pair<A, B> create(A a, B b) {
            return new Pair<A, B>(a, b);
        }
    }
}