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.*;
import java.text.CollationElementIterator;
import java.util.*;

public class B_Yet_Another_Array_Partitioning_Task{
    
    @SuppressWarnings("unchecked")
    public static void main(String[] args) throws Exception{
        int[] nmk = in.readA();
        int n = nmk[0], m = nmk[1], k = nmk[2];
        int[] A = in.readA();

        List<Pair> Ac = new ArrayList<>();

        for(int i = 0 ; i < n ; i++){
            Ac.add(new Pair(i, A[i]));
        }

        Collections.sort(Ac, new Comparator<Pair>(){
            @Override
            public int compare(Pair a, Pair b){
                return a.second - b.second;
            }
        });

        Set<Integer> mxn = new HashSet<>();

        long sum = 0;

        for(int i = 0; i < m*k; i++){
            mxn.add(Ac.get(n-i-1).first);
            sum += Ac.get(n-i-1).second;
        }
        System.out.println(sum );

        int count = 0;
        for(int i = 0; i < n && k > 1; i++){
            if(mxn.contains(i))
                count++;
            if(count == m){
                System.out.print((i+1)+" ");
                k--;
                count = 0;
            }
        }

    }

    static Inputer in;
    static {
        in = new Inputer();
    }

    static class Pair{
        int first, second;
        Pair(int a, int b){
            first = a;
            second = b;
        }
        @Override
        public String toString(){
            return "["+first+" "+second+"]";
        }
    }

    static class Inputer{
        BufferedReader br;
        Inputer(){
            try{
                br = new BufferedReader(new InputStreamReader(System.in));
            }
            catch(Exception e){}
        }
        public int readInt() throws Exception{
            return Integer.parseInt(readLine());
        }
        public long readLong() throws Exception{
            return Long.parseLong(readLine());
        }
        public int[] readA(String delim) throws Exception{
            String[] s = readLine().split(delim);
            int[] A = new int[s.length];
            for(int i = 0; i < s.length; i++)
                A[i] = Integer.parseInt(s[i]);
            return A;
        }
        public int[] readA() throws Exception{
            String[] s = readLine().split("\\s+");
            int[] A = new int[s.length];
            for(int i = 0; i < s.length; i++)
                A[i] = Integer.parseInt(s[i]);
            return A;
        }
        public long[] readLA() throws Exception{
            String[] s = readLine().split("\\s+");
            long[] A = new long[s.length];
            for(int i = 0; i < s.length; i++)
                A[i] = Long.parseLong(s[i]);
            return A;
        }
        public String readLine() throws Exception{
            return br.readLine();
        }
        public int[] copyA(int[] A){
            int[] B = new int[A.length];
            for(int i= 0 ; i < A.length; i++)
                B[i] = A[i];
            return B;
        }
    }
    static void shuffle(int[] A){
        int n = A.length;
        Random rand = new Random();
        for(int t = 0; t < A.length; t++){
            int i1 = rand.nextInt(n);
            int i2 = rand.nextInt(n);
            int tmp = A[i1];
            A[i1] = A[i2];
            A[i2] = tmp;
        }
    }
}