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.io.*;
import java.lang.*;
import java.math.*;
import java.util.Map.*;
public class codeforces {
     static long count =0;
     static boolean f=false;
     static int [] arr;
	static 	PrintWriter pw=new PrintWriter(System.out);
	static void solve(int index , int mask) {
		if(index==arr.length) {
			int sum1=0; int sum2=0;
			for(int i=0;i<arr.length;i++) {
				if((mask & 1<<i)!=0) sum1+=arr[i];
				
			}
			return;
		}
		solve(index+1, mask | 1<<index);
		solve(index+1, mask);
	}
	public static void main(String [] args) throws IOException, InterruptedException {
		Scanner sc=new Scanner(System.in);
		int x=sc.nextInt();
		int y=sc.nextInt();
		int z=sc.nextInt();
		arr=new int [x];
		Integer [] list=new Integer[x];
		TreeMap<Integer, Integer> map=new TreeMap<>();
		for(int i=0;i<x;i++) {
			int k=sc.nextInt();
			arr[i]=k; list[i]=k;
		}
		Long sum=0l;
		Arrays.sort(list);
		for(int i=y*z,j=x;i>0;i--) {
			int k=list[--j];
			sum+=k;
			int b=map.getOrDefault(k,0);
			map.put(k,++b);
		}
		pw.println(sum);
		for(int i=0;i<x;i++) {
			int k=arr[i];
			if(map.containsKey(k)) {
				int b=map.get(k);
				if(b==1) map.remove(k);
				else map.put(k,--b);
				count++;
			}
			if(count==y && --z>0) {pw.print(i+1+" "); count=0; }
		}
		pw.flush();
		pw.close();
	}
	
	
	static class pair implements Comparable<pair>{
		String  name; int x,y,i ;
		public pair(String name , int x) {
			this.name=name; this.x=x;
		}
		
		public pair (int i,int x,int y) {
			this.i=i; this.x=x; this.y=y; 
		}
		public int compareTo(pair o) {
			return x-o.x;
		}
		public int compareTo1(pair o) {
			if(!name.equals(o.name))
				return name.compareTo(o.name);
			return x-o.x;
		}
		public String toString() {
			return i+" "+x+" "+y;
		}
	}

   static class Scanner {
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
	}