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:
/* package whatever; // don't place package name! */

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

/* Name of the class has to be "Main" only if the class is public. */
public class Ideone
{
	static class Pair{
		int value;
		int index;
		Pair(int value,int index)
		{
			this.value=value;
			this.index=index;
		}
	}
	static class sortByIndex implements Comparator<Pair>{
		public int compare(Pair p1,Pair p2)
		{
			return p1.index-p2.index;
		}
	}
	static class sortByValue implements Comparator<Pair>{
		public int compare(Pair p1,Pair p2)
		{
			return (-1)*(p1.value-p2.value);
		}
	}
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner s=new Scanner(System.in);
		int n=s.nextInt();
		int m=s.nextInt();
		int k=s.nextInt();
		
		ArrayList<Integer>al=new ArrayList<>();
		ArrayList<Pair>temp=new ArrayList<Pair>();
		ArrayList<Pair>IndexList=new ArrayList<Pair>();
		for(int i=0;i<n;i++)
		{al.add(s.nextInt());
		temp.add(new Pair(al.get(i),i));
		}
		Collections.sort(temp,new sortByValue());
		long sum=0;
		for(int i=0;i<m*k;i++)
		{
			sum+=temp.get(i).value;
			IndexList.add(temp.get(i));
		}
		System.out.println(sum);
		Collections.sort(IndexList,new sortByIndex());
		// for(int i=0;i<IndexList.size();i++)
		// System.out.println("i "+i+" "+IndexList.get(i).value+" "+IndexList.get(i).index);
		
		
		int t=1;
		for(int i=(m-1);i<IndexList.size();i+=m)
		{
				if(t==k)
				break;
		System.out.print((IndexList.get(i).index+1)+" ");	
			t++;
		}
		
		
	}
}