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.*;
import java.math.*;
import java.lang.*;
 
import static java.lang.Math.*;
 
public class Cf182 implements Runnable 
{
	static class InputReader 
	{
		private InputStream stream;
		private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		public InputReader(InputStream stream) 
		{
			this.stream = stream;
		}
		
		public int read()
		{
			if (numChars==-1) 
				throw new InputMismatchException();
            
			if (curChar >= numChars) 
			{
				curChar = 0;
				try
				{
					numChars = stream.read(buf);
				}
				catch (IOException e)
				{
					throw new InputMismatchException();
				}
                
				if(numChars <= 0)               
					return -1;
			}
			return buf[curChar++];
		}
     
		public String nextLine()
		{
			String str = "";
			try
			{
				str = br.readLine();
			}
			catch (IOException e)
			{
				e.printStackTrace();
			}	
			return str;
		}
		public int nextInt() 
		{
			int c = read();
            
			while(isSpaceChar(c)) 
				c = read();
		
			int sgn = 1;
        
			if (c == '-') 
			{
				sgn = -1;
				c = read();
			}
            
			int res = 0;
			do
			{
				if(c<'0'||c>'9') 
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			}
			while (!isSpaceChar(c)); 
        
			return res * sgn;
		}
        
		public long nextLong() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-')
			{
				sgn = -1;
				c = read();
			}
			long res = 0;
			
			do 
			{
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			}	
			while (!isSpaceChar(c));
				return res * sgn;
		}
		
		public double nextDouble() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-')
			{
				sgn = -1;
				c = read();
			}
			double res = 0;
			while (!isSpaceChar(c) && c != '.') 
			{
				if (c == 'e' || c == 'E')
					return res * Math.pow(10, nextInt());
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			}
			if (c == '.') 
			{
				c = read();
				double m = 1;
				while (!isSpaceChar(c))
				{
					if (c == 'e' || c == 'E')
						return res * Math.pow(10, nextInt());
					if (c < '0' || c > '9')
						throw new InputMismatchException();
					m /= 10;
					res += (c - '0') * m;
					c = read();
				}
			}
			return res * sgn;
		}
    
		public String readString() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do 
			{
				res.appendCodePoint(c);
				c = read();
			} 
			while (!isSpaceChar(c));
            
			return res.toString();
		}
     
		public boolean isSpaceChar(int c) 
		{
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
     
		public String next()
		{
			return readString();
		}
        
		public interface SpaceCharFilter
		{
			public boolean isSpaceChar(int ch);
		}
	}
	public static void main(String args[]) throws Exception 
	{
		new Thread(null, new Cf182(),"Main",1<<27).start();
	}	
	
	public static long gcd(long a, long b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b % a, a); 
	} 
	//  array sorting by colm
public static void sortbyColumn(int arr[][], int col) 
    { 
        
        Arrays.sort(arr, new Comparator<int[]>() { 
            
          @Override
          public int compare(final int[] entry1,  
                             final int[] entry2) { 
  
           
            if (entry1[col] > entry2[col]) 
                return 1; 
            else
                return -1; 
          } 
        });  
    } 
	
	// gcd
  
	public static long findGCD(long arr[], int n) 
	{ 
		long result = arr[0]; 
		for (int i = 1; i < n; i++) 
			result = gcd(arr[i], result); 
		return result; 
	}
	// fibonaci
	static int fib(int n) 
    { 
        int a = 0, b = 1, c; 
        if (n == 0) 
            return a; 
        for (int i = 2; i <= n; i++) 
        { 
            c = a + b; 
            a = b; 
            b = c; 
        } 
        return b; 
    } 
    
// sort a string
    public static String sortString(String inputString) 
    { 
        
        char tempArray[] = inputString.toCharArray(); 
          
         
        Arrays.sort(tempArray); 
          
         
        return new String(tempArray); 
    } 
    // pair function
    
    // list.add(new Pair<>(sc.nextInt(), i + 1));
    //     Collections.sort(list, (a, b) -> Integer.compare(b.first, a.first));
    private static class Pair<F, S> {
 
        private F first;
 
        private S second;
 
        public Pair() {}
 
        public Pair(F first, S second) {
            this.first = first;
            this.second = second;
        }
    }
    private static int next(ArrayList<Integer> arr, int target)  
    {  
        int start = 0, end = arr.size() - 1;  
    
        int ans = -1;  
        while (start <= end) {  
            int mid = (start + end) / 2;  
    
            if (arr.get(mid) <= target) {  
                start = mid + 1;  
            }  
            else {  
                ans = mid;  
                end = mid - 1;  
            }  
        }  
        return ans;  
    }  
    
    public static String LCS(char[] x, char[] y)
    {
      String[][] table= new String[x.length+1][y.length+1];
          for(int i=0;i<=x.length;i++)
          {
              for(int j=0;j<=y.length;j++)
              {
                if(i==0 || j==0)
                    table[i][j] = "";
                else if(x[i-1]==y[j-1])
                    table[i][j] = table[i-1][j-1]+Character.toString(x[i-1]);
                else
                {
                    int p = (table[i-1][j]).length();
                    int q = (table[i][j-1]).length();
                    if(p>q)
                        table[i][j] = table[i-1][j];
                    else
                       table[i][j] = table[i][j-1]; 
                }
            }
              
          }
         return table[x.length][y.length];
    }
    
	public void run()
	{
		InputReader sc = new InputReader(System.in);
		PrintWriter w = new PrintWriter(System.out);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		long ans = 0;
		int a[] = new int[n];
		int temp[] = new int[n];
		for(int i=0;i<n;i++)
		{
		    a[i] = sc.nextInt();
		    temp[i] = a[i];
		}
		Arrays.sort(a);
		int l = n - m*k;
		HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
		for(int i=n-1;i>=l;i--)
	    {
	        ans += a[i];
	        if(map.containsKey(a[i]))
	            map.put(a[i],1+map.get(a[i]));
	        else
	            map.put(a[i],1);
	    }
	    ArrayList<Integer> arr = new ArrayList<Integer>();
	    int count = 0;
	    for(int i=0;i<n;i++)
	    {
	        if(map.containsKey(temp[i]))
	        {
	            int x = map.get(temp[i]);
	            if(x==1)
	                map.remove(temp[i]);
	            else
	                map.put(temp[i],x-1);
	            count++;
	            if(count==m)
	            {
	                count = 0;
	                arr.add(i+1);
	            }
	        }
	        if(arr.size()==k-1)
	            break;
	    }
	    System.out.println(ans);
	    for(int i:arr)
	        System.out.print(i+" ");
	}
}