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 codeforces
			{	
				public static void main(String[] args)
				{
					InputReader in = new InputReader(System.in);
					PrintWriter pw = new PrintWriter(System.out);
					
					int n = in.nextInt();
					int m = in.nextInt();
					int k = in.nextInt();
					
					long[] a = new long[n];
					PriorityQueue<Long> pQueue = new PriorityQueue<>();
					
					for(int i=0;i<n;i++)
					{
						a[i] = in.nextLong();
						
						if(pQueue.size() < m*k)
							pQueue.add(a[i]);
						else
						{
							if(a[i] > pQueue.peek())
							{
								pQueue.poll();
								pQueue.add(a[i]);
							}
						}
					}
					
					int count = 0;
					ArrayList<Integer> list = new ArrayList<>();
					HashMap<Long, Integer> map = new HashMap<>();
					
					for(long p : pQueue)
					{
						if(!map.containsKey(p))
							map.put(p, 1);
						else
							map.put(p, map.get(p) + 1);
					}
					
					for(int i=0;i<n;i++)
					{
						if(map.containsKey(a[i]))
						{
							count++;
							
							if(count == m && list.size() < k-1)
							{
								count = 0;
								list.add(i+1);
							}
							
							map.put(a[i], map.get(a[i])-1);
							
							if(map.get(a[i]) == 0)
								map.remove(a[i]);
						}
					}
					
					long sum = 0;
					
					for(long p : pQueue)
						sum += p;
					
					pw.println(sum);
					
					for(int p : list)
						pw.print(p+" ");
					
					pw.flush();
					pw.close();
				}
				
				public static int BinarySearch_l(long[] a, long k)
				{
					int n = a.length;
					int i = 0, j = n-1;
					int mid = 0;
					
					if(k - a[n-1] > 0)
						return n;
					else
					{
						while(j - i > 0)
						{
							mid = (i+j)/2;
							
							if(k - a[mid] > 0)
								i = mid + 1;
							else
								j = mid;
						}
					}
					
					return i;
				}
				
				public static int BinarySearch_r(long[] a, long k)
				{
					int n = a.length;
					int i = 0, j = n-1;
					int mid = 0;
					
					if(k - a[0] < 0)
						return -1;
					else
					{
						while(j - i > 0)
						{
							mid = (int)Math.ceil((double)(i+j)/2);
							
							if(k - a[mid] >= 0)
								i = mid;
							else
								j = mid - 1;
						}
					}
					
					return i;
				}
				public static long fact(long n, long M)
				{
					long ans = 1;
					
					for(long i=n;i>0;i--)
					{
						ans = (ans*i)%M;
					}
					
					return ans;
				}
				
			/*	public static void DFS(int sourse)
				{
					int count = 0;
					visited[sourse] = true;
					
					for(int u : adj[sourse])
					{
						if(!visited[u])
						{
							DFS(u);
						}
					}
					
					
				}*/
				
				public static ArrayList Divisors(int n)
				{
				    ArrayList<Integer> div = new ArrayList<>();
				    
				    for (int i=1; i<=Math.sqrt(n); i++)
				    {
				        if (n%i == 0)
				        {
				        	div.add(i);
				                
				            if(n/i != i)
				            	div.add(n/i);
				        }
				    }
				    return div;
				}
			
				public static long GCD(long a,long b)
				{
					if(b==0)
						return a;
					else
						return GCD(b,a%b);
				}
				
				public static long LCM(long a,long b)
				{
					return (a*b)/GCD(a, b);
				}
				
				static class pair implements Comparable<pair>
				{
					Integer x, y;
					pair(int x,int y)
					{
						this.x=x;
						this.y=y;
					}
					
					public int compareTo(pair o) {
						int result = x.compareTo(o.x);
						if(result==0)
							result = y.compareTo(o.y);
						
						return result;
					}  
					
					public String toString()
					{
						return x+" "+y;
					}
					
					public boolean equals(Object o)
					{
						if (o instanceof pair)
					    {
							pair p = (pair)o;
							return p.x - x == 0 && p.y - y == 0 ;
						}
						return false;
					}
					
					public int hashCode()
					{
						return new Long(x).hashCode()*31 + new Long(y).hashCode();
					}
				}
				
				static class InputReader 
				{
					 
					private final InputStream stream;
					private final byte[] buf = new byte[8192];
					private int curChar, snumChars;
					private SpaceCharFilter filter;
			 
					public InputReader(InputStream stream) 
					{
						this.stream = stream;
					}
			 
					public int snext() 
					{
						if (snumChars == -1)
							throw new InputMismatchException();
						if (curChar >= snumChars) 
						{
							curChar = 0;
							try 
							{
								snumChars = stream.read(buf);
							} 
							catch (IOException e) 
							{
								throw new InputMismatchException();
							}
							if (snumChars <= 0)
								return -1;
						}
						return buf[curChar++];
					}
			 
					public int nextInt() 
				    {
						int c = snext();
						while (isSpaceChar(c)) 
						{
							c = snext();
						}
						int sgn = 1;
						if (c == '-')
					    {
							sgn = -1;
							c = snext();
						}
						int res = 0;
						do 
						{
							if (c < '0' || c > '9')
								throw new InputMismatchException();
							res *= 10;
							res += c - '0';
							c = snext();
						} while (!isSpaceChar(c));
						return res * sgn;
					}
			 
					public long nextLong()
				    {
						int c = snext();
						while (isSpaceChar(c)) 
						{
							c = snext();
						}
						int sgn = 1;
						if (c == '-') 
						{
							sgn = -1;
							c = snext();
						}
						long res = 0;
						do 
						{
							if (c < '0' || c > '9')
								throw new InputMismatchException();
							res *= 10;
							res += c - '0';
							c = snext();
						} while (!isSpaceChar(c));
						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 String readString()
				    {
						int c = snext();
						while (isSpaceChar(c)) 
						{
							c = snext();
						}
						StringBuilder res = new StringBuilder();
						do 
						{
							res.appendCodePoint(c);
							c = snext();
						} while (!isSpaceChar(c));
						return res.toString();
					}
			 
					public String nextLine() 
					{
						int c = snext();
						while (isSpaceChar(c))
							c = snext();
						StringBuilder res = new StringBuilder();
						do 
						{
							res.appendCodePoint(c);
							c = snext();
						} while (!isEndOfLine(c));
						return res.toString();
					}
			 
					public boolean isSpaceChar(int c) 
					{
						if (filter != null)
							return filter.isSpaceChar(c);
						return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
					}
			 
					private boolean isEndOfLine(int c) 
					{
						return c == '\n' || c == '\r' || c == -1;
					}
			 
					public interface SpaceCharFilter
				    {
						public boolean isSpaceChar(int ch);
					}
				}
				
				static class CodeX {
		        public static void sort(long arr[]) {
		            merge_sort(arr, 0, arr.length - 1);
		        }
		
		        private static void merge_sort(long A[], long start, long end) {
		            if (start < end) {
		                long mid = (start + end) / 2;
		                merge_sort(A, start, mid);
		                merge_sort(A, mid + 1, end);
		                merge(A, start, mid, end);
		            }
		
		        }
		
		        private static void merge(long A[], long start,long mid, long end) {
		            long p = start, q = mid + 1;
		            long Arr[] = new long[(int)(end - start + 1)];
		            long k = 0;
		
		            for (int i = (int)start; i <= end; i++) {
		                if (p > mid)
		                    Arr[(int)k++] = A[(int)q++];
		
		                else if (q > end)
		                    Arr[(int)k++] = A[(int)p++];
		
		                else if (A[(int)p] < A[(int)q])
		                    Arr[(int)k++] = A[(int)p++];
		
		                else
		                    Arr[(int)k++] = A[(int)q++];
		            }
		            for (int i = 0; i < k; i++) {
		                A[(int)start++] = Arr[i];
		            }
		
		        }
		
		    }
			}