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.*;
public class A{
    static PrintWriter out;
    static InputReader in;
    public static void main(String args[]){
        out = new PrintWriter(System.out);
        in = new InputReader();
        new A();
        out.flush(); out.close();
    }   
    class pair{
        int F, S;
        pair(int a, int b){
            F = a; S = b;
        }
    }
    int n, m, k;
    int a[];
    TreeSet<pair> ts = new TreeSet<>((A, B) -> A.F == B.F ? A.S - B.S : A.F - B.F);
    void solveB(){
        n = in.nextInt(); m = in.nextInt(); k = in.nextInt();
        a = new int[n];
        long ans = 0;
        for(int i = 0; i < n; i++){ts.add(new pair(a[i] = in.nextInt(), i)); ans += a[i];}
        while(ts.size() > m * k){
            ans -= ts.pollFirst().F;
        }
        out.println(ans);
        int count = 0;
        for(int i = 0; i < n; i++){
            pair p = ts.ceiling(new pair(a[i], -1));
            if(p != null && p.F == a[i]){
                count++;
                // System.out.println(p.F + " " + p.S);
                ts.remove(p);
            }
            if(count == m && k > 1){
                out.print((i + 1) + " ");
                count = 0; k--;
            }
        }
    }
    A(){
        solveB();
        // solve();
    }
    // final int max = 400010;
    // final int mod = (int)(1e9) + 7;
    // final int maxp = 310;
    // int n, q, tp = 0;
    // int cp[] = new int[maxp];
    // long iv[] = new long[maxp];
    // int order[] = new int[maxp];
    // ArrayList<Integer> p = new ArrayList<>();
    // int f[][] = new int[70][max];
    // long fm[] = new long[max];

    // long mul(long a, long b){
    //     a *= b;
    //     if(a >= mod) a %= mod;
    //     return a;
    // }

    // long rpe(long a, long b){
    //     long ans = 1;
    //     while(b != 0){
    //         ans = mul(ans, a);
    //         a = mul(a, a); b >>= 1;
    //     }return ans;
    // }

    // void seive(){
    //     iv[1] = 1;
    //     for(int i = 2; i < maxp; i++){
    //         iv[i] = rpe(i, mod - 2);
    //     }
    //     for(int i = 1; i < maxp; i++)cp[i] = i;
    //     for(int i = 2; i < maxp; i++){
    //         if(cp[i] != i)continue;
    //         p.add(i); order[i] = ++tp;
    //         for(int j = 2 * i; j < maxp; j += i){
    //             if(cp[j] == j)cp[j] = i;
    //         }
    //     }
    // }

    // void upd(int id, int t, int v){
    //     while(id < max){
    //         f[t][id] += v;
    //         id += id & -id;
    //     }
    // }

    // void updm(int id, long v){
    //     while(id < max){
    //         fm[id] = mul(fm[id], v);
    //         id += id & -id;
    //     }
    // }

    // int qu(int t, int id){
    //     int ans = 0;
    //     while(id > 0){
    //         ans += f[t][id];
    //         id -= id & -id;
    //     }
    //     return ans;
    // }

    // long qm(int id){
    //     long ans = 1;
    //     while(id > 0){
    //         ans = mul(ans, fm[id]);
    //         id -= id & -id;
    //     }return ans;
    // }

    // void add(int l, int r, int x){
    //     updm(l, x); updm(r + 1, iv[x]);
    //     while(x != 1){
    //         int f = cp[x];
    //         upd(l, order[f], 1);
    //         upd(r + 1, order[f], -1);
    //         while(cp[x] == f)x /= cp[x];
    //     }
    // }

    // void solve(){
    //     seive();
    //     // System.out.println(p.size());
    //     n = in.nextInt(); q = in.nextInt();
    //     Arrays.fill(fm, 1);
    //     for(int i = 1; i <= n; i++){
    //         int x = in.nextInt();
    //         add(i, i, x);
    //     }

    //     while(q-- > 0){
    //         if(in.next().trim().compareTo("MULTIPLY") == 0){
    //             int l = in.nextInt(), r = in.nextInt(), x = in.nextInt();
    //             add(l, r, x);
    //         }else{
    //             int l = in.nextInt(), r = in.nextInt();
    //             long a, b;
    //             long tot = mul(a = qm(r), b = rpe(qm(l - 1), mod - 2));
    //             System.out.println(tot + " " + a + " " + b);
    //             boolean rs = false;
    //             for(int f : p){
    //                 int sum = qu(order[f], r) - qu(order[f], l - 1);
    //                 if(sum > 0){
    //                     // System.out.println(sum + " " + f);
    //                     rs = true;
    //                     tot = mul(tot, mul(f - 1, iv[f]));
    //                 }
    //             }
    //             if(!rs) tot = 1;
    //             out.println(tot);
    //         }
    //     }
    // }
    public static class InputReader{
        BufferedReader br;
        StringTokenizer st;
        InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        public int nextInt(){
            return Integer.parseInt(next());
        }
        public long nextLong(){
            return Long.parseLong(next());
        }
        public double nextDouble(){
            return Double.parseDouble(next());
        }
        public String next(){
            while(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){}
            }
            return st.nextToken();
        }
    }
}