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.*;

public class Main {

    private FastScanner scanner = new FastScanner();

    public static void main(String[] args) {
        new Main().solve();
    }

    List<List<Integer>> gr = new ArrayList<>();
    int ans[], res = 0;

    private void solve() {
        int n = scanner.nextInt(), m = scanner.nextInt(), k = scanner.nextInt();

        int a[][] = new int[n][2];

        for (int i = 0; i < n; i++) {
            a[i][0] = scanner.nextInt();
            a[i][1] = i;
        }

        Arrays.sort(a, (o1, o2) -> o2[0] - o1[0]);

        for (int i = k * m; i < n; i++) {
            a[i][0] = Integer.MIN_VALUE;
        }

        Arrays.sort(a, (o1, o2) -> o1[1] - o2[1]);

        int cur = 0;
        long ans = 0;
        List<Integer> r = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            if (a[i][0] != Integer.MIN_VALUE) {
                cur++;
                ans += a[i][0];
            }
            if (cur == m && k > 1) {
                r.add(i + 1);
                cur = 0;
                k--;
            }
        }
        System.out.println(ans);
        for (int i = 0; i < r.size(); i++) {
            System.out.print(r.get(i) + " ");
        }
    }

    public static final <T> void swap(T[] a, int i, int j) {
        T t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    class Pair implements Comparable<Pair> {
        int a, i;

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

        @Override
        public int compareTo(Pair o) {
            return this.a - o.a;
        }
    }

    long gcd(long a, long b) {
        if (b != 0) {
            return gcd(b, a % b);
        }
        return a;
    }

    void initGr(int n, int m) {
        for (int i = 0; i < n; i++) {
            gr.add(new ArrayList<>());
        }

        for (int i = 0; i < m; i++) {
            int u = scanner.nextInt() - 1, v = scanner.nextInt() - 1;
            gr.get(u).add(v);
            gr.get(v).add(u);
        }
    }

    class FastScanner {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }

        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());
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
    }
}