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.math.*;
 
public class Main {
    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(System.out);
        //Scanner sc = new Scanner();
        Reader in = new Reader();
        Main solver = new Main();
        solver.solve(out, in);
        out.flush();
        out.close();
 
    }
 
    //<>
    
    static int maxn = (int)1e5*2;
    static long mod=(int)1e9+7;
    static int n,t,m,k;
    
    void solve(PrintWriter out, Reader in) throws IOException{   
        n = in.nextInt();
        m = in.nextInt();
        k = in.nextInt();
        
        Node[] arr = new Node[n];
        for(int i=0;i<n;i++) arr[i] = new Node(in.nextInt(),i);
        Arrays.sort(arr);
        
        long ans =0;
        ArrayList<Node> list = new ArrayList<>();
        for(int i=n-1;i>=n-m*k;i--){
            ans += arr[i].v;
            list.add(new Node(arr[i].idx,arr[i].v));
        }
        
        Collections.sort(list);
        
        out.println(ans);
        int i=m-1;
        
        int cnt=1;
        while(i<list.size() && cnt<k){
            out.print((list.get(i).v+1)+" ");
            i+=m;
            cnt++;
        }
        out.println();
        
        
    }
    
    static class Node implements Comparable<Node>{
        int idx,v;
        
        Node(int v,int idx){
            this.v = v;
            this.idx = idx;
        }
        
        public int compareTo(Node o){
            return this.v-o.v;
        }
    }
    
    static class Edge implements Comparable<Edge> {
        int u,v,w;
        
        Edge(int u,int v,int w){
            this.u = u;
            this.v = v;
            this.w = w;
        }
        
        
        public int compareTo(Edge o){
            return this.w - o.w;
        }
    }
    
    
    static class Reader {
 
        private InputStream mIs;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
 
        public Reader() {
            this(System.in);
        }
 
        public Reader(InputStream is) {
            mIs = is;
        }
 
        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
 
        }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = mIs.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }
 
        public String nextLine() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        }
 
        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();
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        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 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 boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }
 
        public boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
        }
 
    }
}