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.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.Iterator;
import java.util.Map;
import java.util.Map.Entry;
import java.util.TreeMap;

public class Main {

	private static final String NO = "No";
	private static final String YES = "Yes";
	InputStream is;
	PrintWriter out;
	String INPUT = "";

	private static final long MOD = (long) (1e9 + 7);

	void solve() {
		int T = 1;// ni();
		for (int i = 0; i < T; i++)
			solve(i);
	}

	void solve(int nth) {
		int n = ni();
		int m = ni();
		int k = ni();
		int a[][] = new int[n][2];
		for (int i = 0; i < n; i++) {
			a[i][0] = ni();
			a[i][1] = i + 1;
		}
		Arrays.sort(a, (x, y) -> x[0] - y[0]);
		for (int i = 0; i < n - m * k; i++)
			a[i][1] = n + 100;
		Arrays.sort(a, (x, y) -> x[1] - y[1]);
		long sum = 0;
		for (int i = 0; i < k * m; i++)
			sum += a[i][0];
		out.println(sum);
		for (int i = 1; i < k; i++)
			out.print(a[i * m - 1][1] + " ");
	}

	// a^b
	long power(long a, long b) {
		long x = 1, y = a;
		while (b > 0) {
			if (b % 2 != 0) {
				x = (x * y) % MOD;
			}
			y = (y * y) % MOD;
			b /= 2;
		}
		return x % MOD;
	}

	private long gcd(long a, long b) {
		while (a != 0) {
			long tmp = b % a;
			b = a;
			a = tmp;
		}
		return b;
	}

	void run() throws Exception {
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if (!INPUT.isEmpty())
			tr(System.currentTimeMillis() - s + "ms");
	}

	public static void main(String[] args) throws Exception {
		new Main().run();
	}

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}

	private char[] nc(int n) {
		char[] ret = new char[n];
		for (int i = 0; i < n; i++)
			ret[i] = nc();
		return ret;
	}

	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != '
									// ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n) {
			if (!(isSpaceChar(b)))
				buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private Integer[] na2(int n) {
		Integer[] a = new Integer[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private int[][] na(int n, int m) {
		int[][] a = new int[n][];
		for (int i = 0; i < n; i++)
			a[i] = na(m);
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private Long[] nl2(int n) {
		Long[] a = new Long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}

	private long[] nl(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}

	private long[][] nl(int n, int m) {
		long[][] a = new long[n][];
		for (int i = 0; i < n; i++)
			a[i] = nl(m);
		return a;
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	public static class IterablePermutation implements Iterable<int[]>, Iterator<int[]> {
		int[] a;
		boolean first = true;

		public IterablePermutation(int n) {
			assert n >= 1;
			a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = i;
		}

		public IterablePermutation(int... a) {
			this.a = Arrays.copyOf(a, a.length);
		}

		@Override
		public boolean hasNext() {
			if (first)
				return true;
			int n = a.length;
			int i;
			for (i = n - 2; i >= 0 && a[i] >= a[i + 1]; i--)
				;
			return i != -1;
		}

		@Override
		public int[] next() {
			if (first) {
				first = false;
				return a;
			}
			int n = a.length;
			int i;
			for (i = n - 2; i >= 0 && a[i] >= a[i + 1]; i--)
				;
			assert i != -1;
			int j;
			for (j = i + 1; j < n && a[i] < a[j]; j++)
				;
			int d = a[i];
			a[i] = a[j - 1];
			a[j - 1] = d;
			for (int p = i + 1, q = n - 1; p < q; p++, q--) {
				d = a[p];
				a[p] = a[q];
				a[q] = d;
			}
			return a;
		}

		@Override
		public Iterator<int[]> iterator() {
			return this;
		}
	}

	private static void tr(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

}