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.*;

public class B {
    public static class Pair {
        int a; int b;

        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }

        @Override
        public String toString() {
            return "(" + a + ", " + b + ")";
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt(), m = sc.nextInt(), k = sc.nextInt();
        Pair[] a = new Pair[n];

        for (int i = 0; i < n; i++) {
            a[i] = new Pair(i, sc.nextInt()); // We save the index so we can walk through the m*k best elements in index order later
        }

        // Key: instead of picking and evaluating a partition, instead pick the m*k best elements and then fit the partition
        // around that. The thing is it's always possible to include the m*k best elements in at least one partition.
        // Intuition why: the m first (index-wise) best elements can fit in the first partition, the next m best elements in the next and so on.
        // So annoying how obvious it is once you realize it

        // Find m*k best elements
        List<Pair> best = new ArrayList<>(m*k);

        Arrays.sort(a, new Comparator<Pair>() {
            @Override
            public int compare(Pair o1, Pair o2) {
                return -Integer.compare(o1.b, o2.b); // Want decreasing order
            }
        });

        long sum = 0; // Overflow possible since max is n * 10^9 = 10^5 * 10^9 = 10^14 > 2^32
        for (int i = 0; i < m * k; i++) {
            best.add(a[i]);
            sum += a[i].b;
        }

        // Now sort in index order
        Collections.sort(best, new Comparator<Pair>() {
            @Override
            public int compare(Pair o1, Pair o2) {
                return Integer.compare(o1.a, o2.a);
            }
        });

        System.out.println(sum);

        // Print partitions
        for (int i = 1; i < m * k; i++) {
            if (i % m == 0)
                System.out.print(best.get(i - 1).a + 1 + " ");
        }
        System.out.println();
    }
}