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.LinkedHashMap;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;
import java.util.TreeMap;

public class ArrayPartioningTask {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		int n=scn.nextInt();
		int m=scn.nextInt();
		int k=scn.nextInt();
//		Map<Integer, Integer>lmap = new LinkedHashMap<Integer, Integer>();
//		Map<Integer, Integer> tMap = new TreeMap<Integer, Integer>();
		int z=m*k;
		Integer [][]arr=new Integer[n][2];
		for(int i=0;i<n;i++) {
			int x=scn.nextInt();
			arr[i][0]=x;
			arr[i][1]=i;
//			lmap.put(x, i);
//			tMap.put(x, i);
		}
		Arrays.sort(arr, new Comparator<Integer[]>() {
			@Override
			public int compare(Integer[] o1, Integer[] o2) {
				Integer quantityOne = o1[0];
				Integer quantityTwo = o2[0];
				// reverse sort on quantity
				return quantityOne.compareTo(quantityTwo);
			}
		});		
		//Set<Integer> s=tMap.keySet();
		int cnt=0;
		long sum=0;
		int []y=new int[z];
		for(int i=arr.length-1;i>=0&&cnt<z;i--) {
			//System.out.println(arr[i][0]);
			sum+=arr[i][0];
			y[cnt]=arr[i][1];
			cnt++;
			
		}
		Arrays.sort(y);
		int m1=1,k1=0;
		System.out.println(sum);
		for(int i=0;i<z&&k1<k-1;i++) {
			//System.out.println(m1+" m1");
			if(m1==m) {
				System.out.print((y[i]+1)+" ");
				m1=0;
				k1++;
			}
			m1++;
		}
	}

}