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.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.InputMismatchException;

public class B {
	public static void main(String[] args) {
		FastScannerB sc = new FastScannerB(System.in);

		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		
		int[] arr = new int[n];
		ArrayList<Integer> arr1 = new ArrayList<>(n);
		
		for(int i = 0; i<n; i++){
			arr[i] = sc.nextInt();
			arr1.add(arr[i]);
		}
		
		Collections.sort(arr1, Collections.reverseOrder());
		
		HashMap<Integer, Integer> occurances = new HashMap<>();
		for(int i = 0; i<k*m; i++){
			int val = arr1.get(i);
			if(occurances.containsKey(val)){
				occurances.put(val, occurances.get(val) + 1);
			}
			else{
				occurances.put(val, 1);
			}
		}
		
		ArrayList<Integer> ans = new ArrayList<>();
		int cur_max = 0;
		long sum = 0;
		for(int i = 0; i < n; i++){
			int val = arr[i];
			if(occurances.containsKey(val) && occurances.get(val) > 0){
				sum += val;
				cur_max++;
				occurances.put(val, occurances.get(val) - 1);
				if(cur_max == m){
					ans.add(i);
					cur_max = 0;
				}
			}
		}
		//System.out.println(occu);
		System.out.println(sum);
		
		StringBuilder sb = new StringBuilder();
		for(int i = 0; i<k-1; i++){
			sb.append(ans.get(i) + 1);
			sb.append(' ');
		}
		System.out.println(sb.substring(0, sb.length() - 1));
	}
}

class FastScannerB{
    private InputStream stream;
    private byte[] buf = new byte[1024];
    private int curChar;
    private int numChars;
    private SpaceCharFilter filter;

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

    public int read()
    {
        if (numChars == -1)
            throw new InputMismatchException();
        if (curChar >= numChars)
        {
            curChar = 0;
            try
            {
                numChars = stream.read(buf);
            } catch (IOException e)
            {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }

    public int nextInt()
    {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        int sgn = 1;
        if (c == '-')
        {
            sgn = -1;
            c = read();
        }
        int res = 0;
        do
        {
            if (c < '0' || c > '9')
                throw new InputMismatchException();
            res *= 10;
            res += c - '0';
            c = read();
        } while (!isSpaceChar(c));
        return res * sgn;
    }

    public String next()
    {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuilder res = new StringBuilder();
        do
        {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }
    
    public String nextLine()
    {
        int c = read();
        StringBuilder res = new StringBuilder();
        do
        {
            res.appendCodePoint(c);
            c = read();
        } while (!isLineEndChar(c));
        return res.toString();
    }    
    
    public double nextDouble() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        double res = 0;
        while (!isSpaceChar(c) && c != '.') {
            if (c == 'e' || c == 'E')
                return res * Math.pow(10, nextInt());
            if (c < '0' || c > '9')
                throw new InputMismatchException();
            res *= 10;
            res += c - '0';
            c = read();
        }
        if (c == '.') {
            c = read();
            double m = 1;
            while (!isSpaceChar(c)) {
                if (c == 'e' || c == 'E')
                    return res * Math.pow(10, nextInt());
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                m /= 10;
                res += (c - '0') * m;
                c = read();
            }
        }
        return res * sgn;
    }
    
    public long nextLong() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        int sgn = 1;
        if (c == '-') {
            sgn = -1;
            c = read();
        }
        long res = 0;
        do {
            if (c < '0' || c > '9')
                throw new InputMismatchException();
            res *= 10;
            res += c - '0';
            c = read();
        } while (!isSpaceChar(c));
        return res * sgn;
    }
    
    public boolean isSpaceChar(int c)
    {
        if (filter != null)
            return filter.isSpaceChar(c);
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }
    
    public boolean isLineEndChar(int c)
    {
        if (filter != null)
            return filter.isSpaceChar(c);
        return c == '\n' || c == '\r' || c == -1;
    }

    public interface SpaceCharFilter
    {
        public boolean isSpaceChar(int ch);
    }
}