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.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		int k = in.nextInt();
		ArrayList<p> plist = new ArrayList<p>();
		for (int i = 1; i <= n; i++)
			plist.add(new p(in.nextLong(), i));
		plist.sort(new Comparator<p>() {

			@Override
			public int compare(p p1, p p2) {// 防止long cast到 int之后出现大小问题！
				long delta = p2.value - p1.value;
				if (delta < 0)
					return -1;
				else if (delta == 0)
					return 0;
				else
					return 1;
			}
		});
		List<p> plist1 = plist.subList(0, m * k);
		plist1.sort(new Comparator<p>() {

			@Override
			public int compare(p p1, p p2) {
				return p1.inputOrder - p2.inputOrder;
			}
		});
		long sum = 0;
		for (int i = 0; i < plist1.size(); i++) {
			sum += plist1.get(i).value;
		}
		System.out.println(sum);
		for(int i = m-1; i< plist1.size() && k > 1; i += m,k--) {
			System.out.print(i == m-1?plist1.get(i).inputOrder:" "+plist1.get(i).inputOrder);
		}
		System.out.println();
		in.close();
	}

}

class p {
	long value;
	int inputOrder;

	public p(long value, int inputOrder) {
		super();
		this.value = value;
		this.inputOrder = inputOrder;
	}

	@Override
	public String toString() {
		return "p [value=" + value + ", inputOrder=" + inputOrder + "]";
	}
}