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.*;
import java.math.*;
import java.io.*;
import java.lang.*;

public class Main {
    //public static List<Integer> list = new ArrayList<>();
	public static void main(String[] args) {
        FastReader sc = new FastReader();
        int n = sc.nextInt();
        int m = sc.nextInt();
        int k = sc.nextInt();
        
        //Like my max goal should be to take m*k max values,
        //Now I need to have subarray of max size m;
        //I have the options to break, What my approach is to put all the values I //need in hashmap and then I will traverse the array, whenever I will get m values //which I need for 1 subarray, I will break it there;
        
        int a[] = sc.nextIntArray(n);
        int b[] = a.clone();
        
        //I need to put the values I need in hashmap;
        long needed = (long)(m*k);
        HashMap<Integer,Integer> hm = new HashMap<>();
        long sum =0;
        sort(b);
        for(int i=n-1;i>=0;i--)
        {
            if(needed==0) break;
            hm.put(b[i],hm.getOrDefault(b[i],0)+1);
            sum = sum+b[i];
            needed--;
        }
        
        //So now I have the values I need;
        List<Integer> breaks= new ArrayList<>();
        int subEle = 0; //elements in subarray;
        //I can only add k - 1; breaks;
        for(int i=0;i<n;i++)
        {
            if(hm.containsKey(a[i]) && hm.get(a[i])>=1)
            {
                //I need this element;
                int prevVal = hm.get(a[i]);
                hm.put(a[i],prevVal-1);
                subEle++;
                if(subEle==m)
                {
                    //I have got all m greator elements 
                    //I can break it here;
                    breaks.add((i+1));
                    subEle=0; //starting new subarray;
                    if(breaks.size()==k-1) break;
                }
            }
            else
            {
                //I dont need this;
                //go on;
            }
        }
        System.out.println(sum);
        printList(breaks);
        
        
    }
    
    
    
    
    
    public static void printList(List<Integer> list)
    {
        StringBuffer bf = new StringBuffer("");
        for(int i: list) bf.append(i+" ");
        System.out.println(bf.toString());
    }
    
    
    
    
    
    public static int sum(int a[])
    {
        int sum=0;
        for(int i: a) sum=sum+i;
        return sum;
    }
    
  
    public static int getMax(int a[])
    {
        int m=Integer.MIN_VALUE;
        for(int i: a) m=Math.max(i,m);
        return m;
    }
    
    static class Pair
    {
        int x;
        int y;

        //int i;
        public Pair(int x, int  y)
        {
            this.x=x;
            this.y=y;
        }
    }
    static class Comp implements Comparator<Pair>
    {
        public int compare(Pair p1, Pair p2)
        {
            if(p1.y!=p2.y)
            {
                return p2.y - p1.y;
            }
            return p2.x - p1.x;
        }
    }
    public static int bs(long num[], long x)
    {
        int low=0;
        int high=num.length-1;
        int ans=-1;
        while(low<=high)
        {
            int mid = low + (high-low)/2;
            if(num[mid]==x)
            {
                ans = mid;
                high=mid-1;
            }
            else if(num[mid]<x)
            {
               // ans=mid;
                low=mid+1;
            }
            else
            {
                ans=mid;
                high=mid-1;
            }
        }
        return ans;
    }
    
    
    public static int partition(int arr[], int low, int high) 
    { 
        int pivot = arr[high];  
        int i = (low-1);
        for (int j=low; j<high; j++) 
        { 
            if (arr[j] < pivot) 
            { 
                i++; 
                int temp = arr[i]; 
                arr[i] = arr[j]; 
                arr[j] = temp; 
            } 
        } 
        int temp = arr[i+1]; 
        arr[i+1] = arr[high]; 
        arr[high] = temp; 
  
        return i+1; 
    } 
    public static void sort(int arr[], int low, int high) 
    { 
        if (low < high) 
        { 
            int pi = partition(arr, low, high); 
            sort(arr, low, pi-1); 
            sort(arr, pi+1, high); 
        } 
    } 
    
    
    
    
    
    
    static long gcd(long a,long b)
    {
        if(b==0) return a;
        return gcd(b,a%b);
    }
    
    
    static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
    static void sort(long[] a) {
		ArrayList<Long> l=new ArrayList<>();
		for (long i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	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 (final 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 (final IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		int[] nextIntArray(final int n) {
			final int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = Integer.parseInt(next());
			}
			return a;
		}

	}	
}
//private ArrayList<String[]> action = new ArrayList<String[]>();


//Templates for Comparator and Classes - @jagrit_07
/*
Arrays.sort(newEmployees, new Comparator<Employee>() {
    @Override
    public int compare(Employee emp1, Employee emp2) {
        return emp1.getName().compareTo(emp2.getName());
    }
});
class Pair
	{
		
		long i; //index;
		long l; //left;
		long c; //cost;
		public Pair(long x,long y,long z)
		{
			this.i=x;
			this.l=y;
			this.c=z;
		}
		public String toString()
		{
			return this.i+" "+this.l+" "+this.c;
		}
	}
class Comp implements Comparator<Pair>
	{
		public int compare(Pair p1, Pair p2)
		{
			if(p1.c!=p2.c)
			{
				return (int)(p1.c-p2.c); //sort acc to cost;
			}
			else{
				return (int)(p1.i-p2.i); //sort acc to index;
			}
			
		}
	}
*/

/*
System.out.println( 
      String.format("%.5f", b)); 
  HashMap - Put template - d.put(a1,d.getOrDefault(a1,0)+1);
  for (Map.Entry<String, Integer> entry : map.entrySet()) {
        System.out.println(entry.getKey() + ":" + entry.getValue());
    }
  Deque<String> deque 
            = new LinkedList<String>();  
            
  List<Integer> c[] = new ArrayList[3];          
*/