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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class e {

public static class FastReader {
	BufferedReader br;
	StringTokenizer st;
	//it reads the data about the specified point and divide the data about it ,it is quite fast
	//than using direct 

	public FastReader() {
		br = new BufferedReader(new InputStreamReader(System.in));
	}

	String next() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception r) {
				r.printStackTrace();
			}
		}
		return st.nextToken();
	}

	int nextInt() {
		return Integer.parseInt(next());//converts string to integer
	}

	double nextDouble() {
		return Double.parseDouble(next());
	}

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

	String nextLine() {
		String str = "";
		try {
			str = br.readLine();
		} catch (Exception r) {
			r.printStackTrace();
		}
		return str;
	}
}
static ArrayList<String>list1=new ArrayList<String>();
static void combine(String instr, StringBuffer outstr, int index,int k)
{
	if(outstr.length()==k)
	{
		list1.add(outstr.toString());return;
	}
	if(outstr.toString().length()==0)
	outstr.append(instr.charAt(index));
    for (int i = 0; i < instr.length(); i++)
    {
        outstr.append(instr.charAt(i));
       
        combine(instr, outstr, i + 1,k);
        outstr.deleteCharAt(outstr.length() - 1);
    }
   index++;
} 
static ArrayList<ArrayList<Integer>>l=new ArrayList<>();
static void comb(int n,int k,int ind,ArrayList<Integer>list)
{
	if(k==0)
	{
		l.add(new ArrayList<>(list));

		return;
	}
	
	
	for(int i=ind;i<=n;i++)
	{
		list.add(i);
		comb(n,k-1,ind+1,list);
		
		list.remove(list.size()-1);
		
	}
	
	
	
	
	

}


public static PrintWriter out = new PrintWriter (new BufferedOutputStream(System.out));

	public static void main(String[] args) {
		// TODO Auto-generated method stub

	FastReader in=new FastReader();
	HashMap<Integer,Integer>map=new HashMap<Integer,Integer>();
	HashMap<Integer,Integer>map1=new HashMap<Integer,Integer>();
	ArrayList<Integer>list=new ArrayList<Integer>();
	int n=in.nextInt();int m=in.nextInt();int k=in.nextInt();
	int a[]=new int[n];
	for(int i=0;i<n;i++)
	{
		a[i]=in.nextInt();list.add(a[i]);map1.put(a[i], map1.getOrDefault(a[i], 0)+1);
	}
	Collections.sort(list);
	Collections.reverse(list);long ans=0;
	for(int i=0;i<(m*k);i++)
		{
		map.put(list.get(i), map.getOrDefault(list.get(i), 0)+1);ans+=list.get(i);
		}
	list.clear();int ct=0;
	for(int i=0;i<n;i++)
	{
		if(map.containsKey(a[i])&&map1.get(a[i])<=map.get(a[i]))
		{
			ct++;
			map.put(a[i], map.get(a[i])-1);
		
			if(map.get(a[i])==0)
				map.remove(a[i]);
			
		}
		map1.put(a[i], map1.get(a[i])-1);
		if(map1.get(a[i])==0)
			map1.remove(a[i]);
		if(ct==m&&list.size()<k-1)
		{
			list.add(i+1);ct=0;
		}
		
		
	}
	out.println(ans);
	for(int h:list)
		out.print(h+" ");
	out.close();


	}
}