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 solution {
	
    public static void main (String[] args) throws IOException{
    	FastReader f=new FastReader();
    	int n = f.nextInt();
    	int m = f.nextInt();
    	int k = f.nextInt();
    	int[] a = new int[n];
    	int[] b = new int[n];
    	int[] c = new int[n];
    	HashMap<Integer,ArrayList<Integer>> hm = new HashMap<>();
    	for (int i=0;i<n;i++) {
    		a[i] = f.nextInt();
    		b[i] = a[i];
    		if (hm.get(a[i])==null) hm.put(a[i], new ArrayList<Integer>());
    		hm.get(a[i]).add(i);
    	}
    	
    	Arrays.sort(a);
    	long sum = 0;
    	
    	for (int i=a.length-(m*k);i<a.length;i++) {
    		sum+=a[i];
    		ArrayList<Integer> arr = hm.get(a[i]);
    		c[arr.get(arr.size()-1)] = 1;
    		arr.remove(arr.size()-1);
    	}
    	
    	
    	System.out.println(sum);
    	int prints = 0;
    	for (int i=0;i<c.length;i++) {
    		int count = 0;
    		int j;
    		for (j=i;j<c.length;j++) {
    			if (c[j]==1) count++;
    			if (count==m){
    				System.out.print((j+1)+" ");
    				prints++;
    				if (prints==k-1) return;
    				break;
    			}
    		}
    		i=j;
    	}
    	
    }
    

    public static int binlog( int bits ){
        int log = 0;
        if( ( bits & 0xffff0000 ) != 0 ) { bits >>>= 16; log = 16; }
        if( bits >= 256 ) { bits >>>= 8; log += 8; }
        if( bits >= 16  ) { bits >>>= 4; log += 4; }
        if( bits >= 4   ) { bits >>>= 2; log += 2; }
        return log + ( bits >>> 1 );
    }

    static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
  
}