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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;

/**
 *
 * 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 𝑎𝑙, 𝑎𝑙+1, …, 𝑎𝑟 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 𝑎1,𝑎2,…,𝑎𝑛, 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⋅105, 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 𝑎1,𝑎2,…,𝑎𝑛 (−109≤𝑎𝑖≤109).
 *
 * 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 𝑝1,𝑝2,…,𝑝𝑘−1 (1≤𝑝1<𝑝2<…<𝑝𝑘−1<𝑛) representing the partition of the array, in which:
 *
 * All elements with indices from 1 to 𝑝1 belong to the first subarray.
 * All elements with indices from 𝑝1+1 to 𝑝2 belong to the second subarray.
 * ….
 * All elements with indices from 𝑝𝑘−1+1 to 𝑛 belong to the last, 𝑘-th subarray.
 * If there are several optimal partitions, print any of them.
 *
 * @author fbokovikov
 */
public class Cf_2019_10_02_02 {

    public static void main(String[] args) {
        InputReader reader = new InputReader(System.in);
        int n = reader.nextInt();
        int m = reader.nextInt();
        int k = reader.nextInt();
        long[] array = new long[n];
        for (int i = 0; i < n; i++) {
            array[i] = reader.nextLong();
        }
        partitionArrayWithBeauty(n, m, k, array);
    }

    static void partitionArrayWithBeauty(int n, int m, int k, long[] array) {
        long[] sortedArray = new long[n];
        for (int i = 0; i < n; i++) {
            sortedArray[i] = array[i];
        }
        Arrays.sort(sortedArray);
        HashMap<Long, Integer> maxNumbers = new HashMap<>();
        long beatiesSum = 0;
        for (int i = n - m * k; i < n; i++) {
            maxNumbers.merge(sortedArray[i], 1, (v1, v2) -> v1 + v2);
            beatiesSum += sortedArray[i];
        }
        int[] partitionKeys = new int[k - 1];
        int pk = 0;
        int totalMaxElemsCount = 0;
        for(int i = 0; i < n; i++) {
            Integer updatedCount = maxNumbers.computeIfPresent(array[i], (key, value) -> value - 1);
            if (updatedCount != null && updatedCount >= 0) {
                totalMaxElemsCount++;
            }
            if (totalMaxElemsCount == m && pk < (k - 1)) {
                partitionKeys[pk] = i + 1;
                pk++;
                totalMaxElemsCount = 0;
            }
        }
        System.out.println(beatiesSum);
        for (int i = 0; i < k - 1; i++) {
            System.out.print(partitionKeys[i] + " ");
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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