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.awt.Point;
import java.io.*;
public class Solutions {

	static FastScanner scr=new FastScanner();
	static PrintStream out;
	static ArrayList<ArrayList<Integer>>list=new ArrayList<ArrayList<Integer>>();
	public static void main(String[] args) {
		out = new PrintStream(System.out);
		int n=read();
		int m=read();
		int k=read();
		TreeSet<Point>ts=new TreeSet<>(new Comparator<Point>() {
//			@Override
			public int compare(Point a,Point b) {
				 int diff=b.x-a.x;
				 return (diff==0)?(a.y-b.y):diff;
			}
		});
		for(int i=1;i<=n;i++) {
			int a=read();
			ts.add(new Point(a,i));
		}
		Iterator<Point>it=ts.iterator();
		
		TreeSet<Integer>index=new TreeSet<Integer>();
		long sum=0;
		for(int i=0;i<k*m;i++) {
			Point p=it.next();
			index.add(p.y);
			sum+=p.x;
		}
		System.out.println(sum);
		Iterator<Integer>itr=index.iterator();
		for(int l=1;l<k;l++) {
			int last=-1;
			for(int i=0;i<m;i++) {
				last=itr.next();
			}
			System.out.print(last+" ");
		}
		
	}
	static  int read() {
		return scr.nextInt();
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	static int MAX=Integer.MAX_VALUE;
	static int MIN=Integer.MIN_VALUE;
}class Triplet{
	int first;
	int second;
	int third;
	Triplet(int fisrt,int second,int third){
		this.first=fisrt;
		this.second=second;
		this.third=third;
	}
}
class pair{
	long x=0;
	int y=0;
	pair(long first,int second){
		this.x=first;
		this.y=second;
	}
}