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.lang.*;
import java.io.*;
import java.math.BigDecimal;
public class R538B {
    public static void main (String[] args) throws java.lang.Exception {
        InputReader in = new InputReader(System.in);
        PrintWriter w = new PrintWriter(System.out);
        int n = in.nextInt(), m = in.nextInt(), k = in.nextInt();
        int[] a = new int[n], b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
            b[i] = a[i];
        }
        Arrays.sort(b);
        HashMap<Integer, Integer> hm = new HashMap<>();
        long ans = 0;
        for (int i = 0; i < m * k; i++) {
            if (!hm.containsKey(b[n - 1 - i]))
                hm.put(b[n - 1 - i], 0);

            hm.put(b[n - 1 - i], hm.get(b[n - 1 - i]) + 1);
            ans += b[n - 1 - i];
        }
        w.println(ans);
        int count = 0;
        for (int i = 0; i < n && count / m < k; i++) {
            if (hm.containsKey(a[i])) {
                count++;
                if (count % m == 0 && count / m < k)
                    w.print((i + 1) + " ");
                hm.put(a[i], hm.get(a[i]) - 1);
                if (hm.get(a[i]) == 0)
                    hm.remove(a[i]);
            }

        }
        w.close();
    }
    static boolean nextPermutation(int[] a) {
        int n = a.length;
        int ptr = n - 1;
        while (ptr > 0 && a[ptr - 1] >= a[ptr]) {
            ptr--;
        }

        for (int i = ptr, j = n - 1; i < j; i++, j--) {
            int tmp = a[i];
            a[i] = a[j];
            a[j] = tmp;
        }

        if (ptr == 0) {
            return false;
        }

        for (int i = ptr;; i++) {
            if (a[ptr - 1] < a[i]) {
                int tmp = a[ptr - 1];
                a[ptr - 1] = a[i];
                a[i] = tmp;
                return true;
            }
        }
    }
    static int fibo(int n) {
        if (n == 1 || n == 2)
            return 1;
        return fibo(n - 1) + fibo(n - 2);
    }
    /*static int fibo(int n) {
        if (n == 0)
            return 0;
        if (n == 1 || n == 2)
            return f[n] = 1;
        if (f[n] != 0)
            return f[n];
        if (n % 2 == 0) {
            int k = (n) / 2;
            return f[n] = (fibo(k - 1) * 2 + fibo(k)) * fibo(k);
        } else {
            int k = (n + 1) / 2;
            return f[n] = fibo(k - 1) * fibo(k - 1) + fibo(k) * fibo(k);
        }
    }*/
    static class Edge implements Comparable<Edge> {
        int from, to, cost;
        Edge(int from, int to, int cost) {
            this.from = from;
            this.to = to;
            this.cost = cost;
        }
        public int compareTo(Edge O) {
            return this.cost - O.cost;
        }
    }
    static String swap(String s, int i, int j) {
        char[] ca = s.toCharArray();
        char temp = ca[i];
        ca[i] = ca[j];
        ca[j] = temp;
        return new String(ca);
    }
    static void findMissExtra(int[] a) {
        HashSet<Integer> hs = new HashSet<>();
        int n = a.length, extra = 0, miss = 0;
        for (int i = 0; i < n; i++) {
            if (hs.contains(a[i]))
                extra = a[i];
            hs.add(a[i]);

        }
        for (int i = 1; i <= n; i++) {
            if (!hs.contains(i)) {
                miss = i;
                break;
            }
        }
        System.out.println(miss + ", " + extra);
    }
    static void find(int m, int s) {
        if (s == 0 || s > m * 9) {
            System.out.println("Impossible");
            return ;
        }
        s -= 1;
        int[] num = new int[m];
        for (int i = m - 1; i > 0; i--) {
            int temp = Math.min(s, 9);
            num[i] = temp;
            s -= temp;
        }
        num[0] += s + 1;
        for (int x : num)
            System.out.print(x);
    }
    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;

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

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

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

        public void skip(int x) {
            while (x-- > 0)
                read();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public String nextString() {
            return next();
        }

        public String next() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuffer res = new StringBuffer();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isSpaceChar(c));

            return res.toString();
        }

        public String nextLine() {
            StringBuffer buf = new StringBuffer();
            int c = read();
            while (c != '\n' && c != -1) {
                if (c != '\r')
                    buf.appendCodePoint(c);
                c = read();
            }
            return buf.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 int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }
        public long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
        public boolean hasNext() {
            int value;
            while (isSpaceChar(value = peek()) && value != -1)
                read();
            return value != -1;
        }

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

    }
}