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.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;
import java.util.TreeSet;


public class Main {

	static class Task {
		
		int NN = 200005;
		int MOD = 1000000007;;
		int INF = 2000000000;
		long INFINITY = 2000000000000000000L;
		
		Long [] a ;
		
		class Pair {
			Long value;
			Integer index;
			public Pair(Long v, Integer i) {
				this.value = v;
				this.index = i;
			}
		}
		
		public void solve(InputReader in, PrintWriter out) {
			int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
			TreeSet<Pair> Q = new TreeSet<>(
					(p1, p2) -> p1.value.equals(p2.value)
					? p1.index.compareTo(p2.index)
					: p2.value.compareTo(p1.value));
			a = new Long[n];
			for(int i=0;i<n;++i) {
				a[i] = in.nextLong();
				Q.add(new Pair(a[i], i));
			}
			long ans = 0;
			Map<Integer, Boolean> map = new HashMap<>();
			for(int i=n-1;i>(n-1)-(k*m);--i) {
				Pair top = Q.pollFirst();
				ans += top.value;
				map.put(top.index, true);
			}
			out.println(ans);
			int total = 0;
			int cnt = 0;
			for(int i=0;i<n;++i) {
				if(map.containsKey(i))
					++cnt;
				if(cnt == m) {
					++total;
					out.print((1 + i) + " ");					
					cnt = 0;
					if(total == k - 1)
						break;
				}
			}
		}
		
	}
	
	static void prepareIO(boolean isFileIO) {
		//long t1 = System.currentTimeMillis();
		Task solver = new Task();
		// Standard IO
		if(!isFileIO) {
			InputStream inputStream = System.in;
	        OutputStream outputStream = System.out;
	        InputReader in = new InputReader(inputStream);
	        PrintWriter out = new PrintWriter(outputStream);
	        solver.solve(in, out);
	        //out.println("time(s): " + (1.0*(System.currentTimeMillis()-t1))/1000.0);
	        out.close();
		}
        // File IO
		else {
			String IPfilePath = System.getProperty("user.home") + "/Downloads/ip.in";
	        String OPfilePath = System.getProperty("user.home") + "/Downloads/op.out";
	        InputReader fin = new InputReader(IPfilePath);
	        PrintWriter fout = null;
	        try {
				fout = new PrintWriter(new File(OPfilePath));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
	        solver.solve(fin, fout);
	        //fout.println("time(s): " + (1.0*(System.currentTimeMillis()-t1))/1000.0);
	        fout.close();
		}
	}
	
	public static void main(String[] args) {
        prepareIO(false);
	}
	
	static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        
        public InputReader(String filePath) {
        	File file = new File(filePath);
            try {
				reader = new BufferedReader(new FileReader(file));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
            tokenizer = null;
        }
        
        public String nextLine() {
        	String str = "";
        	try {
				str = reader.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
        	return str;
        }
        
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

}