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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class appy {

	// Driver program
	public static void main(String[] args) {
		FastReader scn = new FastReader();
		int n = scn.nextInt();
		int m = scn.nextInt();
		int k = scn.nextInt();
		int a[] = new int[n];
		ArrayList<Pair> set = new ArrayList<>();
		for (int i = 0; i < n; ++i) {
			a[i] = scn.nextInt();
			set.add(new Pair(a[i], i));
		}
		long sum = 0;
		Collections.sort(set, new mySort1()); // sorts in decreasing order
		int i;
		for (i = 0; i < m * k; ++i) {
			sum += set.get(i).x;
		}
		int times = n - (m * k);
		while (times-- > 0) {
			set.remove(set.size()-1);
		}
		System.out.println(sum);
		Collections.sort(set, new mySort());
		int kk = k - 1;
		for (i = m - 1; i < set.size() && kk != 0;) {
			System.out.print(set.get(i).y + 1 + " ");
			kk--;
			i += m;
		}
	}

	private static void delete(int n, ArrayList<Pair> set) {
		for (int i = 0; i < set.size(); ++i) {
			if (set.get(i).x == n) {
				set.remove(i);
				return;
			}
		}
	}

	public 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;
		}
	}
}

class mySort implements Comparator<Pair> {
	@Override
	public int compare(Pair o1, Pair o2) {
		return o1.y - o2.y;
	}
}

class mySort1 implements Comparator<Pair> {
	@Override
	public int compare(Pair o1, Pair o2) {
		if (o1.x == o2.x) {
			return o1.y - o2.y;
		} else {
			return o2.x - o1.x;
		}
	}
}

class Pair {
	int x;
	int y;

	Pair(int x, int y) {
		this.x = x;
		this.y = y;
	}
}