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.BigInteger;
import java.text.*;
public class Main {
    static long mod = 1000_000_007;
    static long mod1 = 998244353;
    static boolean fileIO = false;
    static boolean memory = true;
    static FastScanner f;
    static PrintWriter pw;
    static double eps = (double)1e-6;
    static FileWriter fw;
    static long oo = Long.MAX_VALUE;
 
    // N = 1 or N = max ?
    // longs vs. ints ?
    // max / min ?

    public static void solve() throws Exception {
        int n = f.ni() , m = f.ni() , k = f.ni();

        TreeMap<Integer , TreeSet<Integer>> mp = new TreeMap<>();
        
        for(int i = 0; i < n; ++i) {
            int x = f.ni();
            if(!mp.containsKey(x)) mp.put(x , new TreeSet<>());
            mp.get(x).add(i + 1);
        }
        
        Integer inds[] = new Integer[m * k];
        long sum = 0;

        for (int i = 0; i < k * m; ++i){
            int l = mp.lastKey();
            sum += l;
            inds[i] = mp.get(l).pollLast();
            if(mp.get(l).size() == 0) mp.remove(l);
        }
        
        Arrays.sort(inds);
        int idx = -1;
        pn(sum);

        for(int i = 1; i <= k - 1; ++i)
            p((inds[idx += m]) + " ");
    }      
 
    public static void main(String[] args)throws Exception {
        if(memory) new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();System.exit(1);}}}, "", 1 << 28).start();
        else new Main().run();
    }
        
/******************************END OF MAIN PROGRAM*******************************************/
    void run()throws Exception {
        if (System.getProperty("ONLINE_JUDGE") == null) {
            f = new FastScanner("");
            pw = new PrintWriter(System.out);
            fw = new FileWriter("!out.txt");
        }
        else {
            f = new FastScanner();
            pw = new PrintWriter(System.out);
            fw = new FileWriter("!out.txt");
        }
        //
        int t = 1;
        int tt = 1;

        while(t --> 0) {
            //p("Case #" + (tt++) + ": ");
            //fw.write("\n");
            solve();
        }
        
        pw.flush(); 
        pw.close(); 
    }
 
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner(String str) throws Exception {
            try {
                br = new BufferedReader(new FileReader("!a.txt"));
            }
            catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        public String next()throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
 
        public int ni() throws IOException {return Integer.parseInt(next());}
 
        public long nl() throws IOException {return Long.parseLong(next());}
 
        public String nextLine() throws IOException {return br.readLine();}
 
        public double nd() throws IOException {return Double.parseDouble(next());}
 
    }
 
    public static void pn(Object... o) {for(int i = 0; i < o.length; ++i) pw.print(o[i] + (i + 1 < o.length ? " ": "\n"));}
    public static void p(Object... o) {for(int i = 0; i < o.length; ++i) pw.print(o[i] + (i + 1 < o.length ? " " : ""));}
    public static void pni(Object... o) {for(Object obj : o) pw.print(obj + " "); pw.println(); pw.flush();}
    public static int gcd(int a,int b){if(b==0)return a;else{return gcd(b,a%b);}}
    public static long gcd(long a,long b){if(b==0l)return a;else{return gcd(b,a%b);}}
    public static long lcm(long a,long b){return (a*b/gcd(a,b));}
    public static long pow(long a,long b){long res=1;while(b>0){if((b&1)==1)res=res*a;b>>=1;a=a*a;}return res;}
    public static int pow(int a,int b){int res=1;while(b>0){if((b&1)==1)res=res*a;b>>=1;a=a*a;}return res;}
    public static long mpow(long a,long b, long m){long res=1;while(b>0){if((b&1)==1)res=((res%m)*(a%m))%m;b>>=1;a=((a%m)*(a%m))%m;}return res;}
    public static long mul(long a , long b , long mod){return ((a%mod)*(b%mod)%mod);}
    public static long adp(long a , long b){return ((a%mod)+(b%mod)%mod);}
    public static int dig(long a){int cnt=0;while(a>0){a/=10;++cnt;}return Math.max(1,cnt);}
    public static int dig(int a){int cnt=0;while(a>0){a/=10;++cnt;}return Math.max(1,cnt);}
    public static boolean isPrime(long n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(long i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
    public static boolean isPrime(int n){if(n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
    public static HashSet<Long> factors(long n){HashSet<Long> hs=new HashSet<Long>();for(long i=1;i<=(long)Math.sqrt(n);i++){if(n%i==0){hs.add(i);hs.add(n/i);}}return hs;}
    public static HashSet<Integer> factors(int n){HashSet<Integer> hs=new HashSet<Integer>();for(int i=1;i<=(int)Math.sqrt(n);i++){if(n%i==0){hs.add(i);hs.add(n/i);}}return hs;}
    public static HashSet<Long> pf(long n){HashSet<Long> ff=factors(n);HashSet<Long> ans=new HashSet<Long>();for(Long i:ff)if(isPrime(i))ans.add(i);return ans;}
    public static HashSet<Integer> pf(int n){HashSet<Integer> ff=factors(n);HashSet<Integer> ans=new HashSet<Integer>();for(Integer i:ff)if(isPrime(i))ans.add(i);return ans;}
    public static int gnv(char c){return Character.getNumericValue(c);}
    public static void sort(int[] a){ArrayList<Integer> l=new ArrayList<>();for(int i:a)l.add(i);Collections.sort(l);for(int i=0;i<a.length;++i)a[i]=l.get(i);}
    public static void sort(long[] a){ArrayList<Long> l=new ArrayList<>();for(long i:a)l.add(i);Collections.sort(l);for(int i=0;i<a.length;++i)a[i]=l.get(i);}
    public static void sort(ArrayList<Integer> a){Collections.sort(a);}
}