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.util.PriorityQueue;
import java.io.IOException;
import java.io.Reader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.io.BufferedReader;
import java.util.Comparator;
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;
        MyInput in = new MyInput(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, MyInput in, PrintWriter out) {
            PriorityQueue<int[]> que = new PriorityQueue<>(Comparator.comparingInt(x -> -x[0]));
            int n = in.nextInt();
            int m = in.nextInt();
            int k = in.nextInt();
            int[] a = in.nextIntArray(n);
            for (int i = 0; i < n; i++) {
                que.add(new int[]{a[i], i});
            }
            long sum = 0;
            int[] cnt = new int[n];
            for (int i = 0; i < m * k; i++) {
                int[] p = que.poll();
                sum += p[0];
                cnt[p[1]]++;
            }
            out.println(sum);
            List<Integer> p = new ArrayList<>();
            for (int i = 0, cur = 0; i < n; i++) {
                cur += cnt[i];
                if (cur == m && p.size() < k - 1) {
                    p.add(i + 1);
                    cur = 0;
                }
            }
            printList(out, p);
        }

        static void printList(PrintWriter out, List<Integer> p) {
            for (int i = 0; i < p.size(); i++) {
                out.print(p.get(i) + (i == p.size() - 1 ? "\n" : " "));
            }
        }

    }

    static class MyInput {
        private final BufferedReader in;
        private static int pos;
        private static int readLen;
        private static final char[] buffer = new char[1024 * 8];
        private static char[] str = new char[500 * 8 * 2];
        private static boolean[] isDigit = new boolean[256];
        private static boolean[] isSpace = new boolean[256];
        private static boolean[] isLineSep = new boolean[256];

        static {
            for (int i = 0; i < 10; i++) {
                isDigit['0' + i] = true;
            }
            isDigit['-'] = true;
            isSpace[' '] = isSpace['\r'] = isSpace['\n'] = isSpace['\t'] = true;
            isLineSep['\r'] = isLineSep['\n'] = true;
        }

        public MyInput(InputStream is) {
            in = new BufferedReader(new InputStreamReader(is));
        }

        public int read() {
            if (pos >= readLen) {
                pos = 0;
                try {
                    readLen = in.read(buffer);
                } catch (IOException e) {
                    throw new RuntimeException();
                }
                if (readLen <= 0) {
                    throw new MyInput.EndOfFileRuntimeException();
                }
            }
            return buffer[pos++];
        }

        public int nextInt() {
            int len = 0;
            str[len++] = nextChar();
            len = reads(len, isSpace);
            int i = 0;
            int ret = 0;
            if (str[0] == '-') {
                i = 1;
            }
            for (; i < len; i++) ret = ret * 10 + str[i] - '0';
            if (str[0] == '-') {
                ret = -ret;
            }
            return ret;
        }

        public char nextChar() {
            while (true) {
                final int c = read();
                if (!isSpace[c]) {
                    return (char) c;
                }
            }
        }

        int reads(int len, boolean[] accept) {
            try {
                while (true) {
                    final int c = read();
                    if (accept[c]) {
                        break;
                    }
                    if (str.length == len) {
                        char[] rep = new char[str.length * 3 / 2];
                        System.arraycopy(str, 0, rep, 0, str.length);
                        str = rep;
                    }
                    str[len++] = (char) c;
                }
            } catch (MyInput.EndOfFileRuntimeException e) {
            }
            return len;
        }

        public int[] nextIntArray(final int n) {
            final int[] res = new int[n];
            for (int i = 0; i < n; i++) {
                res[i] = nextInt();
            }
            return res;
        }

        static class EndOfFileRuntimeException extends RuntimeException {
        }

    }
}