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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.util.*;
import java.math.*;

import org.omg.CORBA.Any;


public class template {

	
	
	static int n=0, m=0,k=0;
	public static void main(String[] args) throws IOException {
       Reader scn=new Reader();
       n=scn.nextInt(); m=scn.nextInt(); k=scn.nextInt();
       
       int []a=new int[n];

        for(int i=0; i<n; i++)
        	a[i]=scn.nextInt();
        
       PriorityQueue<pair> pq=new PriorityQueue<>(); //pair type to store the value-a[i] and index i
       
       for(int i=0; i<n; i++){  //loop to store the highest m*k values
    	   
    	    if(i<m*k)
    	    	pq.add(new pair(a[i],i));
    	    
    	    else{
    	    	 if(a[i]>pq.peek().val){
    	    		  pq.remove();
    	    		  pq.add(new pair(a[i], i));
    	    	 }
    	    }
       }
       
       ArrayList<Integer> p=new ArrayList<>();
       
       while(pq.size()!=0)    //adding the indices of the top m*k value into an arraylist
    	   p.add(pq.remove().idx);
       
       Collections.sort(p);  // now the top m*k value are sorted acc to their index
       
       
       long cnt=1, max=0;
       StringBuilder sb=new StringBuilder();
       
       for(int i=0; i<p.size(); i++){
    	   
    	   if(cnt % m == 0 && cnt<=m*k-m){    // making cuts at separation of m for k-1 subarrays
    		   sb.append(p.get(i)+1);
    		   sb.append(" ");
    	   }
    	   max+=a[p.get(i)];  // finding the sum of highest m*k values which is the total beauty
    	   
    	   cnt++;
       }
       
       
       System.out.println(max);
       System.out.println(sb);
	}







	// _________________________TEMPLATE_____________________________________________________________

	private static class pair implements Comparable<pair> {
		int val;
		int idx;

		pair(int l, int b) {
			val = l;
			idx= b;
		}

		@Override
		public int compareTo(pair o) {
			
			return this.val-o.val;
		}

		
	}

	


	public static void printarray(long[] dp1) {
		for (long i : dp1)
			System.out.print(i + " ");
		System.out.println();
	}
	
	public static void inputarray(long []a, int n,Reader scn) throws IOException{
		for(int i=0; i<n; i++)
			a[i]=scn.nextLong();
	}

	private static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[100000 + 1]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}
	}

}