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:
// package june2020;

import java.util.*;
import java.lang.*;
import java.math.*;
import java.io.*;

public class cf_subs {
	static FastReader scn = new FastReader();
	static OutputStream out = new BufferedOutputStream(System.out);
//	@SuppressWarnings("unchecked")
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		int t = 1;
		while(t-->0) {
			int n =scn.nextInt(),m=scn.nextInt(),k=scn.nextInt();
//			int[] a = new int[n];
			pair[] p = new pair[n];
			for(int i=0;i<n;i++) {
				int sv=scn.nextInt();
				p[i]=new pair(sv,i);
			}
			Arrays.sort(p);
//			int[] b=mergesort(a,0,n-1);
			long ans=0;
			ArrayList<Integer> inds = new ArrayList<>();
			for(int i=n-1;i>=n-m*k;i--) {
				ans+=p[i].x;
				inds.add(p[i].y);
			}
			
			out.write((ans+"\n").getBytes());
			Collections.sort(inds);
			ArrayList<Integer> divs = new ArrayList<>();
			for(int i=0;i<k-1;i++) {
				divs.add(inds.get((i+1)*m-1));
			}
			for(int i=0;i<divs.size();i++) {
				out.write((divs.get(i)+1+" ").getBytes());
			}
			out.write("\n".getBytes());
		}
		out.close();
	}
	static class pair implements Comparable<pair>{
		int x,y;
		public pair(int a,int b) {
			this.x=a;
			this.y=b;
		}
		public int compareTo(pair o) {
			if(this.x==o.x) {
				return Integer.compare(this.y, o.y);
			}
			return Integer.compare(this.x, o.x);
		}
	}
	public static int[] merge(int[] one,int[] two){
		int[] res = new int[one.length+two.length];
		int i = 0;
		int j = 0;
		int k = 0;
		
		while(i<one.length&&j<two.length){
			if(one[i]<two[j]){
				res[k] = one[i];
				i++;
				k++;
			}
			else{
				res[k] = two[j];
				j++;
				k++;
			}
		}
		
		if(i==one.length){
			while(j<two.length){
				res[k] = two[j];
				j++;
				k++;
			}
		}
		else{
			while(i<one.length){
				res[k] = one[i];
				k++;
				i++;
			}
		}
		return res;
	}
	public static int[] mergesort(int[] arr, int l, int r){
		if(l==r){
			int[] br = new int[1];
			br[0] = arr[l];
			return br;
		}
		int mid = (l+r)/2;
		int[] fh = mergesort(arr,l,mid);
		int[] sh = mergesort(arr,mid+1,r);
		
		return merge(fh,sh);
	}
//	static int dfs(int idx) {
//		
//	}
	static <T> int upperBound(List<T> ls, T x) throws RuntimeException {
		if (ls.size() == 0)
			return -1;
		if (ls.get(0) instanceof Integer) {
			return ~Collections.binarySearch(ls, x, (t1, t2) -> ((Integer) t1).compareTo((Integer) t2) > 0 ? 1 : -1);
		} else if (ls.get(0) instanceof Long) {
			return ~Collections.binarySearch(ls, x, (t1, t2) -> ((Long) t1).compareTo((Long) t2) > 0 ? 1 : -1);
		} else if (ls.get(0) instanceof Double) {
			return ~Collections.binarySearch(ls, x, (t1, t2) -> ((Double) t1).compareTo((Double) t2) > 0 ? 1 : -1);
		} else {
			System.err.println(
					String.format("%s:Arey Maa chudi padi hai", Thread.currentThread().getStackTrace()[1].getMethodName()));
			throw new RuntimeException();
		}
	}
	static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }

}