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.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class BBBBBBBB {
	static Pair ar[];
	static int m;
	static int k;
	static int n;
	public static void work()
	{
		long ans=0L;
		Arrays.sort(ar,new sortone());
	
		
		for(int i=0;i<(k*m);i++)
		{
			ans+=ar[i].val;
			//if(!(i==(k*m)-1))
			ar[i].selec=1;
			
		}
		Arrays.sort(ar,new sorttwo());
		System.out.println(ans);
		int curr=0;
		int z=0;
		for(int i=0;i<n;i++)
		{
			if(ar[i].selec==1)
			{
				curr++;
			}
			if(curr>=m && z<k)
				
			{
				
				if(!(z==k-1))
				System.out.print(i+1+" ");
				curr=0;
				z++;
			}
			
		}
		
		
	}
	public static void main(String args[])
	{
		Scanner scan=new Scanner(System.in);
		n=scan.nextInt();
		m=scan.nextInt();
		k=scan.nextInt();
		ar=new Pair[n];
		for(int i=0;i<n;i++)
		{
			ar[i]=new Pair(scan.nextInt(),i);
		}
		work();
	}
}
class Pair{
	int val=0;
	int inx=0;
	int selec=0;
	Pair(int a,int b)
	{
		val=a;
		inx=b;
	}
}
class sortone implements Comparator<Pair>
{

	@Override
	public int compare(Pair a, Pair b) {
		if(a.val<b.val)
		{
			return 1;
		}else if(b.val<a.val){
			return -1;
		}else{
			return 0;
		}
	}
	
	
}
class sorttwo implements Comparator<Pair>
{

	@Override
	public int compare(Pair o1, Pair o2) {
		if(o1.inx>o2.inx)
		{
			return 1;
		}
		else if(o1.inx<o2.inx)
		{
			return -1;
		}
		else{
			return 0;
		}
	}
	
}