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.io.*;
import java.math.*;

public class Main {
	static class Solution {
		public void solve(InputReader in, OutputWriter out) {
			int N = in.nextInt();
			int m = in.nextInt();
			int k = in.nextInt();
			int[] a = new int[N];
			Integer[] idx = new Integer[N];
			for (int i = 0; i < N; i++) {
				a[i] = in.nextInt();
				idx[i] = i;
			}
			Arrays.sort(idx, (e1, e2) -> a[e2] - a[e1]);
			int elem = k * m;
			long total = 0;
			int[] cuts = new int[elem];
			for (int i = 0; i < elem; i++) {
				total += a[idx[i]];
				cuts[i] = idx[i];
			}
			Arrays.sort(cuts);
			System.out.println(total);
			for (int i = 0; i < elem; i++) {
				if (i % m == m - 1 && i != elem - 1) {
					System.out.print((cuts[i] + 1) + " ");
				}
			}
		}
	}

	public static void main(String[] args) {
		InputReader in = new InputReader(System.in);
		OutputWriter out = new OutputWriter(System.out);
		Solution sol = new Solution();
		sol.solve(in, out);
		out.close();
	}
	
	static class InputReader {
		BufferedReader br;
		StringTokenizer st;
		InputReader(InputStream in) {br = new BufferedReader(new InputStreamReader(in));}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() {return Double.parseDouble(next());}
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {st = new StringTokenizer(br.readLine());}
				catch (IOException e) {e.printStackTrace();}
			}
			return st.nextToken();
		}
		String nextLine() {
			String str = "";
			try {str = br.readLine();}
			catch (IOException e) {e.printStackTrace();}
			return str;
		}
	}

	static class OutputWriter {
		PrintWriter pw;
		OutputWriter(OutputStream out) {pw = new PrintWriter(new BufferedOutputStream(out), true);}
		void printf(String s, Object... objects) {print(String.format(s, objects));}
		void println(Object... objects) {print(objects); pw.println();}
		void close() {pw.close();}
		void print(Object... objects) {
			for (int i = 0; i < objects.length; i++) {
				if (i != 0) pw.print(' ');
				pw.print(objects[i]);
			}
		}
	}
}