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.util.*;
public class d {
    public static void main(String[] args) throws IOException {
        Scanner s = new Scanner(System.in);
    //BufferedReader s=new BufferedReader(new InputStreamReader(System.in));
        //       String[] st=s.readLine().trim().split("\\s+");
        //      a=Integer.parseInt(st[i]);
        //  String[] st1=s.readLine().trim().split("\\s+");
        StringBuilder sb = new StringBuilder();
        StringBuilder sb1 = new StringBuilder();
        int n=s.nextInt();
        int m=s.nextInt();
        int k=s.nextInt();
        int[] a=new int[n];
        Student[] st=new Student[n];
        for(int i=0;i<n;i++){
            a[i]=s.nextInt();
            st[i]=new Student(a[i],i);

        }int[] val=new int[n];
        Arrays.sort(st,new Sortbyroll());
        long sum=0;
        for(int i=0;i<k*m;i++){
            val[(int)st[i].r]=1;sum+=st[i].l;
        }int count=0;int c=0;
        for(int i=0;i<n;i++){
            if(val[i]==1){
                count++;
            }
            if(count==m&&c<k-1){c++;
                sb.append((i+1)+" ");count=0;
            }
        }
        System.out.println(sum);
        System.out.println(sb.toString());
    }
    static int[] vis;
    public static int countSetBits(int n)
    {
        return (BitsSetTable256[n & 0xff]
                + BitsSetTable256[(n >> 8) & 0xff]
                + BitsSetTable256[(n >> 16) & 0xff]
                + BitsSetTable256[n >> 24]);
    }
    static int[] BitsSetTable256 ;
    public static void initialize(int n)
    {
        BitsSetTable256[0] = 0;
        for (int i = 0; i <=Math.pow(2,n); i++) {
            BitsSetTable256[i] = (i & 1) + BitsSetTable256[i / 2];
        }
    }
    static HashMap<Integer,Integer>[] val;//static int[] vis;static int y;
    static boolean dfs(int x ,int i,ArrayList<Integer>[] adj){
      vis[i]=1;
     // if(x==1) System.out.print(i+" ");
      if(adj[i]==null) return false;
      for(int j:adj[i]){
        //  if(x==1) System.out.print(j+" ");
          if(j==x) return true;
          if(vis[j]==0) return dfs(x,j,adj);
      }return false;
    }

    static  void computeLPSArray(String pat, int M, int lps[]) {
        int len = 0;
        int i = 1;
        lps[0] = 0;
        while (i < M) {
            if (pat.charAt(i) == pat.charAt(len)) {
                len++;
                lps[i] = len;
                i++;
            }
            else
            {
                if (len != 0) {
                    len = lps[len - 1];
                }
                else
                {
                    lps[i] = len;
                    i++;
                }
            }
        }
    }
    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;
    }
    static long powerwithmod(long  x, long y, int p) {
        long  res = 1;
        x = x % p;
        while (y > 0) {
            if ((y & 1) == 1)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }
    static long  powerwithoutmod(long x, int y) {
        long  temp;
        if( y == 0)
            return 1;
        temp = powerwithoutmod(x, y/2);
        if (y%2 == 0)
            return temp*temp;
        else
        {
            if(y > 0)
                return x * temp * temp;
            else
                return (temp * temp) / x;
        }
    }
    static void fracion(double x) {
        String a = "" + x;
        String spilts[] = a.split("\\."); // split using decimal
        int b = spilts[1].length(); // find the decimal length
        int denominator = (int) Math.pow(10, b); // calculate the denominator
        int numerator = (int) (x * denominator); // calculate the nerumrator Ex
        // 1.2*10 = 12
        int gcd = (int) gcd((long) numerator, denominator); // Find the greatest common
        // divisor bw them
        String fraction = "" + numerator / gcd + "/" + denominator / gcd;
        // System.out.println((denominator/gcd));
        long x1 = modInverse(denominator / gcd, 998244353);
        //  System.out.println(x1);
        System.out.println((((numerator / gcd) % 998244353 * (x1 % 998244353)) % 998244353));
    }
    static int bfs(int i1, ArrayList<Integer>[] h, int[] vis, int n,int val1) {
        Queue<Integer> q = new LinkedList<Integer>();
        q.add(i1);Queue<Integer> aq=new LinkedList<Integer>();
        aq.add(0);
        while(!q.isEmpty()){
            int i=q.poll();
            int val=aq.poll();
            if(i==n){
                return val;
            }
            if(h[i]!=null){
                for(Integer j:h[i]){
                    if(vis[j]==0){
                        q.add(j);vis[j]=1;
                        aq.add(val+1);}
                }
            }
        }return -1;
    }
    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
    static long modInverse(long a, int m)
    {
        return (powerwithmod(a, m - 2, m));
    }
    static int MAXN=100001;
    static int[] spf=new int[MAXN];
    static void sieve() {
        spf[1] = 1;
        for (int i=2; i<MAXN; i++)
            spf[i] = i;
        for (int i=4; i<MAXN; i+=2)
            spf[i] = 2;
        for (int i=3; i*i<MAXN; i++)
        {
            if (spf[i] == i)
            {
                for (int j=i*i; j<MAXN; j+=i)

                    if (spf[j]==j)
                        spf[j] = i;
            }
        }
    }
    static ArrayList<Integer> getFactorizationUsingSeive(int x) {
        ArrayList<Integer> ret = new ArrayList<Integer>();
        while (x != 1)
        {
            ret.add(spf[x]);
            x = x / spf[x];
        }
        return ret;
    }
   /*static long[] fac = new long[MAXN+1];
     static void calculatefac(int mod){
        for (int i = 1 ;i <= MAXN; i++)
            fac[i] = fac[i-1] * i % mod;
     }
      static long nCrModPFermat(int n, int r, int mod) {
         if (r == 0)
            return 1;
        fac[0] = 1;
        return (fac[n]* modInverse(fac[r], mod)
                % mod * modInverse(fac[n-r], mod)
                                    % mod) % mod;
    } */}
class Student {
    long  l;long r;int x;
    public Student(long l, long r) {
        this.l = l;
        this.r = r;this.x=x;
    }

    public String toString()
    {
        return this.l+" ";
    }
}
class Sortbyroll implements Comparator<Student>
{

    public int compare(Student a, Student b){
      if(a.l<b.l) return 1;
      else if(a.l==b.l){
          if(a.r==b.r){
              return 0;
          }
          if(a.r<b.r) return -1;
          return 1;}
      return -1;  }
}