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.math.BigInteger;
import java.util.*;

import javax.sound.midi.Synthesizer;
public class tr {
static int n,m,s;
static PrintWriter out;
static long [] v,t,p,res,d;
static double [] a;
static ArrayList<Integer>[]ad;
static boolean []vis,vis2;
public static void main(String[] args) throws IOException{
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
	    n=sc.nextInt();
	    int m=sc.nextInt();
	    int k =sc.nextInt();
	    int k1=k;
	    Integer []a=new Integer[n];
	    for(int i=0;i<n;i++)
	    	a[i]=sc.nextInt();
	    Integer []p=a.clone();
	    Arrays.sort(p);
	    int le=m*k;
	    HashMap <Integer,Integer> hm=new HashMap();
	    long max=0;
	    for(int i=n-1;i>=0 && le-->0;i--) {
	    	max+=p[i];
	    	if(hm.containsKey(p[i]))
	    		hm.put(p[i],hm.get(p[i])+1);
	    	else
				hm.put(p[i],1);
	    }
	   //System.out.println(max+" "+hm);
	    long ans=0;
	    
	    Queue <Integer> help=new LinkedList();int i=0;
	    while(i<n) {
	    	int c=0;
	    	k--;
	    	while( k*m<n-i+1 &&c<m&&i<n) {
	    		if(hm.containsKey(a[i])) {
	    			c++;
	    			ans+=a[i];
	    			if(hm.get(a[i])==1)
	    				hm.remove(a[i]);
	    			else
	    				hm.put(a[i],hm.get(a[i])-1);
	    
	    		}
	    		i++;
	    	}
	    //	System.out.println(hm);
	    	help.add(i);
	    }
	    //System.out.println(ans+" "+help);
	    out.println(ans);
	    for(int u=0;u<k1-1;u++)
	    	out.print(help.remove()+" ");
 		out.flush();
	}
static int c;
static void dfs(int u) {
	vis[u]=true;
	for(int v:ad[u])
		if(!vis[v])
			dfs(v);
}
static void dfs2(int u) {
	vis2[u]=true; c++;
	for(int v:ad[u])
		if(!vis2[v])
			dfs2(v);
}
static class node implements Comparable<node>{
	int node;
	int reach;
	node(int n,int r){
		node=n;
		reach=r;
	}
	public int compareTo(node o) {
		return o.reach-reach;
	}
	public String toString() {
		return node+" "+reach;
	}
}
	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);
		}
	}
}