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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

import java.util.Arrays;

import java.util.Comparator;

import java.util.StringTokenizer;

public class Main {
	static class Point{
		int x,y;
		public Point(int x,int y) {
			this.x=x;
			this.y=y;
		}
		
	}
	
	public static void main(String[] args) {	
		FastScanner sc=new FastScanner();
		int n=sc.nextInt();
		int m=sc.nextInt();
		int k=sc.nextInt();
		int arr[]=new int[n];
		for(int i=0;i<n;i++) {
			arr[i]=sc.nextInt();
		}
		Point[] brr=new Point[n]; 
		for(int i=0;i<n;i++) {
			brr[i]=new Point(arr[i], i);
		}
		Arrays.sort(brr,new Comparator<Point>() {

			@Override
			public int compare(Point o1, Point o2) {
				// TODO Auto-generated method stub
				if(o2.x==o1.x)return o2.y-o1.y;
				return o2.x-o1.x;
			}
		});
		long total=0;
		boolean[] flag=new boolean[n];
		for(int i=0;i<m*k;i++) {
			total+=brr[i].x;
			flag[brr[i].y]=true;
		}
		
		System.out.println(total);
		int t=0;
		int count=0;
		for(int i=0;i<n;i++) {
			if(flag[i])count++;
			if(count==m) {
				System.out.print((i+1)+" ");
				t++;
				count=0;
			}
			if(t==k-1)break;
		}
		System.out.println();
		
	}
	
	
	
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");

		public String next() {
			while (!st.hasMoreElements())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

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

		public int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) {
				a[i]=nextInt();
			}
			return a;
		}

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

	}

}