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.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStreamWriter;
import java.math.BigInteger;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Random;
import java.util.TreeSet;

public final class CF_538_D2_B
{





	static boolean verb=true;
	static void log(Object X){if (verb) System.err.println(X);}
	static void log(Object[] X){if (verb) {for (Object U:X) System.err.print(U+" ");System.err.println("");}}
	static void log(int[] X){if (verb) {for (int U:X) System.err.print(U+" ");System.err.println("");}}
	static void log(int[] X,int L){if (verb) {for (int i=0;i<L;i++) System.err.print(X[i]+" ");System.err.println("");}}
	static void log(long[] X){if (verb) {for (long U:X) System.err.print(U+" ");System.err.println("");}}

	static void logWln(Object X){if (verb) System.err.print(X);}
	static void info(Object o){	System.out.println(o);}
	static void output(Object o){outputWln(""+o+"\n");	}
	static void outputWln(Object o){try {out.write(""+ o);} catch (Exception e) {}}

	static long powerMod(long b,long e,long m){
		long x=1;
		while (e>0) {
			if (e%2==1)
				x=(b*x)%m;
			b=(b*b)%m;
			e=e/2;
		}
		return x; 
	}


	static long pgcd(long a,long b){
		if (b>a)
			return pgcd(b,a);
		while (b>0){
			long c=b;
			b=a%b;
			a=c;
		}
		return a;
	}

	static class Composite implements Comparable<Composite>{
		int v;
		int i;
		public int compareTo(Composite X) {
		if (v<X.v)
			return 1;
		if (v>X.v)
			return -1;
			return i-X.i;
		}
		public Composite(int v, int i) {
			this.v = v;
			this.i = i;
		}
		
		
	}
	

	static long mod=1000000007;

	// Global vars
	static BufferedWriter out;
	static InputReader reader;


	
	
	static void process() throws Exception {

		out = new BufferedWriter(new OutputStreamWriter(System.out));
		reader = new InputReader(System.in);

		int n=reader.readInt();
		int m=reader.readInt();
		int k=reader.readInt();
		Composite[] ar=new Composite[n];
		int[] a=new int[n];
		
		for (int i=0;i<n;i++) {
			a[i]=reader.readInt();
			ar[i]=new Composite(a[i],i);
		}
		int[] marked=new int[n];
		
		Arrays.sort(ar);
		long sum=0;
		for (int i=0;i<m*k;i++) {
			marked[ar[i].i]=1;
			sum+=ar[i].v;
		}
		//log(marked);
		int cnt=0;
		int kk=0;
		output(sum);
		log("m:"+m);
		for (int i=0;i<n;i++) {
			if (marked[i]==1) {
				cnt++;
				if (cnt==m) {
					if (kk<k-1) 
					outputWln((i+1)+" ");
					kk++;
					
					cnt=0;
				}
			}
		}
		
		
		try {
			out.close();
		} catch (Exception e) {
		}

	}

	public static void main(String[] args) throws Exception {
		process();

	}

	static final class InputReader {
		private final InputStream stream;
		private final byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		private int read() throws IOException {
			if (curChar >= numChars) {
				curChar = 0;
				numChars = stream.read(buf);
				if (numChars <= 0) {
					return -1;
				}
			}
			return buf[curChar++];
		}

		public final String readString() throws IOException {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			StringBuilder res = new StringBuilder();
			do {
				res.append((char) c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public final int readInt() throws IOException {
			int c = read();
			boolean neg = false;
			while (isSpaceChar(c)) {
				c = read();
			}
			char d = (char) c;
			// log("d:"+d);
			if (d == '-') {
				neg = true;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			// log("res:"+res);
			if (neg)
				return -res;
			return res;

		}

		public final long readLong() throws IOException {
			int c = read();
			boolean neg = false;
			while (isSpaceChar(c)) {
				c = read();
			}
			char d = (char) c;
			// log("d:"+d);
			if (d == '-') {
				neg = true;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			// log("res:"+res);
			if (neg)
				return -res;
			return res;

		}

		private boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
	}

}