Problem: An array b is called to be a subarray of a if it forms a continuous subsequence of a, that is, if it is equal to a_l, a_{l + 1}, …, a_r for some l, r.

Suppose m is some known constant. For any array, having m or more elements, let's define it's beauty as the sum of m largest elements of that array. For example: 

  * For array x = [4, 3, 1, 5, 2] and m = 3, the 3 largest elements of x are 5, 4 and 3, so the beauty of x is 5 + 4 + 3 = 12.
  * For array x = [10, 10, 10] and m = 2, the beauty of x is 10 + 10 = 20.



You are given an array a_1, a_2, …, a_n, the value of the said constant m and an integer k. Your need to split the array a into exactly k subarrays such that:

  * Each element from a belongs to exactly one subarray.
  * Each subarray has at least m elements.
  * The sum of all beauties of k subarrays is maximum possible.

Input

The first line contains three integers n, m and k (2 ≤ n ≤ 2 ⋅ 10^5, 1 ≤ m, 2 ≤ k, m ⋅ k ≤ n) — the number of elements in a, the constant m in the definition of beauty and the number of subarrays to split to.

The second line contains n integers a_1, a_2, …, a_n (-10^9 ≤ a_i ≤ 10^9).

Output

In the first line, print the maximum possible sum of the beauties of the subarrays in the optimal partition.

In the second line, print k-1 integers p_1, p_2, …, p_{k-1} (1 ≤ p_1 < p_2 < … < p_{k-1} < n) representing the partition of the array, in which:

  * All elements with indices from 1 to p_1 belong to the first subarray.
  * All elements with indices from p_1 + 1 to p_2 belong to the second subarray.
  * ….
  * All elements with indices from p_{k-1} + 1 to n belong to the last, k-th subarray.



If there are several optimal partitions, print any of them.

Examples

Input


9 2 3
5 2 5 2 4 1 1 3 2


Output


21
3 5 

Input


6 1 4
4 1 3 2 2 3


Output


12
1 3 5 

Input


2 1 2
-1000000000 1000000000


Output


0
1 

Note

In the first example, one of the optimal partitions is [5, 2, 5], [2, 4], [1, 1, 3, 2].

  * The beauty of the subarray [5, 2, 5] is 5 + 5 = 10. 
  * The beauty of the subarray [2, 4] is 2 + 4 = 6. 
  * The beauty of the subarray [1, 1, 3, 2] is 3 + 2 = 5. 



The sum of their beauties is 10 + 6 + 5 = 21.

In the second example, one optimal partition is [4], [1, 3], [2, 2], [3].

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.OutputStream;
import java.util.Iterator;
import java.util.PriorityQueue;
import java.util.Collection;
import java.util.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Objects;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BYetAnotherArrayPartitioningTask solver = new BYetAnotherArrayPartitioningTask();
        solver.solve(1, in, out);
        out.close();
    }

    static class BYetAnotherArrayPartitioningTask {
        public void solve(int testNumber, FastReader in, PrintWriter out) {
            Debug debug = new Debug();
            int n = in.nextInt();
            int m = in.nextInt();
            int K = in.nextInt();
            long[] a = in.nextLongArray(n);

            PriorityQueue<BYetAnotherArrayPartitioningTask.Pair> priorityQueue = new PriorityQueue<>();
            for (int i = 0; i < n; ++i) priorityQueue.add(new BYetAnotherArrayPartitioningTask.Pair(i, a[i]));
            boolean[] marked = new boolean[n];
            int rem = m * K;
            while (rem != 0) {
                BYetAnotherArrayPartitioningTask.Pair cur = priorityQueue.poll();
                marked[cur.ind] = true;
                --rem;
            }
            long ans = 0;
            ArrayList<Integer> ansList = new ArrayList<>();
            int remK = K;
            for (int i = 0; i < n; ) {
                int j = i;
                int ct = 0;
                PriorityQueue<Long> now = new PriorityQueue<>(Collections.reverseOrder());
                while (j < n && ct < m) {
                    if (marked[j]) ++ct;
                    now.add(a[j]);
                    ++j;
                }
                --remK;
                while (remK == 0 && j < n) now.add(a[j++]);
                if (j < n) ansList.add(j);
                for (int p = 0; p < m; ++p) ans += now.poll();
                i = j;
            }
            out.println(ans);
            ArrayUtils.printArray(out, ansList);
        }

        static class Pair implements Comparable<BYetAnotherArrayPartitioningTask.Pair> {
            int ind;
            long value;

            public Pair(int i, long v) {
                ind = i;
                value = v;
            }

            public boolean equals(Object o) {
                if (this == o) return true;
                if (o == null || getClass() != o.getClass()) return false;
                BYetAnotherArrayPartitioningTask.Pair pair = (BYetAnotherArrayPartitioningTask.Pair) o;
                return ind == pair.ind &&
                        value == pair.value;
            }

            public int hashCode() {
                return Objects.hash(ind, value);
            }

            public int compareTo(BYetAnotherArrayPartitioningTask.Pair o) {
                return Long.compare(o.value, value);
            }

        }

    }

    static class Debug {
        PrintWriter out;
        boolean oj;
        long timeBegin;
        Runtime runtime;

        public Debug(PrintWriter out) {
            oj = System.getProperty("ONLINE_JUDGE") != null;
            this.out = out;
            this.timeBegin = System.currentTimeMillis();
            this.runtime = Runtime.getRuntime();
        }

        public Debug() {
            oj = System.getProperty("ONLINE_JUDGE") != null;
            OutputStream outputStream = System.err;
            this.out = new PrintWriter(outputStream);
            this.timeBegin = System.currentTimeMillis();
            this.runtime = Runtime.getRuntime();
        }

        public void finalize() {
            out.flush();
            out.close();
        }

    }

    static class ArrayUtils {
        public static <T> void printArray(PrintWriter out, Collection<T> array) {
            if (array.size() == 0) return;
            int size = array.size();
            Iterator<T> iterator = array.iterator();
            int ct = 0;
            for (; ct < size && iterator.hasNext(); ct++) {
                if (ct != 0) out.print(" ");
                out.print(iterator.next());
            }
            out.println();
        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[8192];
        private int curChar;
        private int pnumChars;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        private int pread() {
            if (pnumChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= pnumChars) {
                curChar = 0;
                try {
                    pnumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (pnumChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = pread();
            while (isSpaceChar(c))
                c = pread();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = pread();
            }
            int res = 0;
            do {
                if (c == ',') {
                    c = pread();
                }
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = pread();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public long nextLong() {
            int c = pread();
            while (isSpaceChar(c))
                c = pread();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = pread();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = pread();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public long[] nextLongArray(int n) {
            long[] array = new long[n];
            for (int i = 0; i < n; i++) {
                array[i] = nextLong();
            }
            return array;
        }

        private boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

    }
}