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.*;
import java.util.*;


public class codeforces2 {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter pw = new PrintWriter(System.out);
//        Scanner scan = new Scanner(System.in);
        int tc = 1;
        for (int rep = 0; rep < tc; rep++) {
            int N = sc.ni();
            int M = sc.ni();
            int K = sc.ni();
            int[] arr = sc.intArray(N);
            pw.println( solve(N, M, K, arr));
        }
        pw.close();
    }
    public static String solve(int N, int M, int K, int[] arr) {
        int[][] ref = new int[N][2];
        for (int i = 0; i <ref.length; i++) {
            ref[i] = new int[] {arr[i], i};
            
        }
        Arrays.sort(ref, new Comparator<int[]>() {
            public int compare(int[] a, int[] b) {
                return -1 * Integer.compare(a[0], b[0]);
            }
        });
//        System.out.println(Arrays.deepToString(ref));
        List<Integer> list = new ArrayList();
        long ret = 0;
        for (int i = 0; i < M*K; i++) {
            ret += ref[i][0];
            list.add(ref[i][1]);
        }
        Collections.sort(list);
//        System.out.println(list.toString());
        StringBuilder sb = new StringBuilder();
        sb.append(ret);
        sb.append("\n");
        
        for (int i = 0; i < K-1; i++) {
            sb.append((list.get((i+1)*M-1) + 1) + " ");
        }
        return sb.toString();
        
    }
      
        
    static long pow(long num, long exp, long mod){
        long ans=1;
        for(int i=1;i<=exp;i++){
            ans=(ans*num)%mod;
        }
        return ans;
    }
    static boolean isPrime(int n)
    {
 
        // Check if number is less than
        // equal to 1
        if (n <= 1)
            return false;
 
        // Check if number is 2
        else if (n == 2)
            return true;
 
        // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;
 
        // If not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) 
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b; 
        return gcd(b % a, a); 
    }
     
    // method to return LCM of two numbers
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
    public static void sort(int[] arr) {
        Random rgen = new Random();
        for (int i = 0; i < arr.length; i++) {
            int r = rgen.nextInt(arr.length);
            int temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr);
    }
 
    public static void sort(long[] arr) {
        Random rgen = new Random();
        for (int i = 0; i < arr.length; i++) {
            int r = rgen.nextInt(arr.length);
            long temp = arr[i];
            arr[i] = arr[r];
            arr[r] = temp;
        }
        Arrays.sort(arr);
    }
    /* */
    //printing methods
    /* */
    //WOW!
    /* */
    public static void printArr(PrintWriter pw, int[] arr) {
        StringBuilder sb = new StringBuilder();
        for (int x : arr) {
            sb.append(x + "");
        }
        sb.setLength(sb.length() - 1);
        pw.println(sb.toString());
    }
    public static void printArr2d(PrintWriter pw, int[][] arr) {
        StringBuilder sb = new StringBuilder();
        for (int[] row : arr) {
            for (int x : row) {
                sb.append(x + " ");
            }
            sb.setLength(sb.length() - 1);
            sb.append("\n");
        }
        sb.setLength(sb.length() - 1);
        pw.println(sb.toString());
    }
}
class FastScanner {
    BufferedReader br;
    StringTokenizer st;

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

    String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    int ni() {
        return Integer.parseInt(next());
    }

    int[] intArray(int N) {
        int[] ret = new int[N];
        for (int i = 0; i < N; i++)
            ret[i] = ni();
        return ret;
    }

    long nl() {
        return Long.parseLong(next());
    }

    long[] longArray(int N) {
        long[] ret = new long[N];
        for (int i = 0; i < N; i++)
            ret[i] = nl();
        return ret;
    }

    double nd() {
        return Double.parseDouble(next());
    }

    String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
}
class UnionFind {
    int size;
    private int[] id;
    public UnionFind(int size) {
        this.size = size;
        id = new int[size];
        for (int i = 0; i < id.length; i++) {
            id[i] = i;
        }
    }
    public int find(int a) {
        if (id[a] != a) {
            id[a] = find(id[a]);
        }
        return id[a];
    }
    public void union(int a, int b) {
        int r1 = find(a);
        int r2 = find(b);
        if (r1 == r2) return;
        size--;
        id[r1] = r2;
    }
    @Override
    public String toString() {
        return Arrays.toString(id);
    }
}