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.awt.Point;
import java.math.BigInteger;

public class split {
	
	static Comparator<Point> comp = new Comparator<Point>() {
		
		public int compare(Point a, Point b) {
			return Integer.compare(b.x, a.x);
		}
		
	};
	
	public static void main(String[] args) throws Exception {
		FastIO sc = new FastIO(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		
		int[] a = new int[n];
		for(int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		Point[] b = new Point[n];
		for(int i = 0; i < n; i++) {
			b[i] = new Point(a[i], i);
		}
		boolean[] yeah = new boolean[n];
		
		long total = 0;
		Arrays.sort(b, comp);
		for(int i = 0; i < m*k; i++) {
			yeah[b[i].y] = true;
			total += b[i].x;
		}
		
		ArrayList<Integer> indices = new ArrayList<Integer>();
		
		int counter = 0;
		for(int i = 0; i < n; i++) {
			if(yeah[i]) counter++;
			if(counter == m) {
				indices.add(i+1);
				counter = 0;
			}
		}
		
		pw.println(total);
		for(int i = 0; i < indices.size()-1; i++) {
			pw.print(indices.get(i));
			if(i != indices.size()-2) pw.print(" ");
		}
		
		pw.println();
		pw.close();
		
		
	}

	static class FastIO {

		//Is your Fast I/O being bad?

		InputStream dis;
		byte[] buffer = new byte[1 << 17];
		int pointer = 0;

		public FastIO(String fileName) throws Exception {
			dis = new FileInputStream(fileName);
		}

		public FastIO(InputStream is) throws Exception {
			dis = is;
		}

		int nextInt() throws Exception {
			int ret = 0;

			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			boolean negative = false;
			if (b == '-') {
				negative = true;
				b = nextByte();
			}
			while (b >= '0' && b <= '9') {
				ret = 10 * ret + b - '0';
				b = nextByte();
			}

			return (negative) ? -ret : ret;
		}

		long nextLong() throws Exception {
			long ret = 0;

			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			boolean negative = false;
			if (b == '-') {
				negative = true;
				b = nextByte();
			}
			while (b >= '0' && b <= '9') {
				ret = 10 * ret + b - '0';
				b = nextByte();
			}

			return (negative) ? -ret : ret;
		}

		byte nextByte() throws Exception {
			if (pointer == buffer.length) {
				dis.read(buffer, 0, buffer.length);
				pointer = 0;
			}
			return buffer[pointer++];
		}

		String next() throws Exception {
			StringBuffer ret = new StringBuffer();

			byte b;
			do {
				b = nextByte();
			} while (b <= ' ');
			while (b > ' ') {
				ret.appendCodePoint(b);
				b = nextByte();
			}

			return ret.toString();
		}

	}
}