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.*;
import java.io.*;
public class B {

	public static void main(String[] args) throws IOException{
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in)); 
		StringTokenizer st = new StringTokenizer(f.readLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		ArrayList<Long> arr = new ArrayList<Long>(n);
		ArrayList<Long> sortedarr = new ArrayList<Long>(n);
		st = new StringTokenizer(f.readLine());
		for(int i = 0; i < n; i++) {
			long a = Long.parseLong(st.nextToken());
			arr.add(a);
			sortedarr.add(a);
		}
		Collections.sort(sortedarr);
		int last = (n-(m*k));
		long arrsum = 0L;
		HashMap<Long, Integer> elems = new HashMap<Long, Integer>();
		for(int i = last; i < n; i++) {
			arrsum+=sortedarr.get(i);
			if(!elems.containsKey(sortedarr.get(i)))
				elems.put(sortedarr.get(i), 0);
			elems.put(sortedarr.get(i), elems.get(sortedarr.get(i))+1);			
		}
		ArrayList<Integer> partitions = new ArrayList<Integer>();
		int currElements = 0;
		for(int i = 0; i < n; i++) {
			if(elems.containsKey(arr.get(i))) {
				currElements++;
				elems.put(arr.get(i), elems.get(arr.get(i))-1);
				if(elems.get(arr.get(i)) == 0) elems.remove(arr.get(i));
			}
			if(currElements == m) {
				currElements = 0;
				partitions.add(i+1);
			}
		}
		System.out.println(arrsum);
		for(int i = 0; i < partitions.size()-2; i++) {
			System.out.print(partitions.get(i));
			System.out.print(" ");
		}
		System.out.print(partitions.get(partitions.size()-2));
		System.out.println();
		
		

	}

}