Helen works in Metropolis airport. She is responsible for creating a departure schedule. There are n flights that must depart today, the i-th of them is planned to depart at the i-th minute of the day.

Metropolis airport is the main transport hub of Metropolia, so it is difficult to keep the schedule intact. This is exactly the case today: because of technical issues, no flights were able to depart during the first k minutes of the day, so now the new departure schedule must be created.

All n scheduled flights must now depart at different minutes between (k + 1)-th and (k + n)-th, inclusive. However, it's not mandatory for the flights to depart in the same order they were initially scheduled to do so — their order in the new schedule can be different. There is only one restriction: no flight is allowed to depart earlier than it was supposed to depart in the initial schedule.

Helen knows that each minute of delay of the i-th flight costs airport ci burles. Help her find the order for flights to depart in the new schedule that minimizes the total cost for the airport.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 300 000), here n is the number of flights, and k is the number of minutes in the beginning of the day that the flights did not depart.

The second line contains n integers c1, c2, ..., cn (1 ≤ ci ≤ 107), here ci is the cost of delaying the i-th flight for one minute.

Output

The first line must contain the minimum possible total cost of delaying the flights.

The second line must contain n different integers t1, t2, ..., tn (k + 1 ≤ ti ≤ k + n), here ti is the minute when the i-th flight must depart. If there are several optimal schedules, print any of them.

Example

Input

5 2
4 2 1 10 2


Output

20
3 6 7 4 5 

Note

Let us consider sample test. If Helen just moves all flights 2 minutes later preserving the order, the total cost of delaying the flights would be (3 - 1)·4 + (4 - 2)·2 + (5 - 3)·1 + (6 - 4)·10 + (7 - 5)·2 = 38 burles. 

However, the better schedule is shown in the sample answer, its cost is (3 - 1)·4 + (6 - 2)·2 + (7 - 3)·1 + (4 - 4)·10 + (5 - 5)·2 = 20 burles.

import java.util.*;
import java.io.*;
public class Solution {
	static FastScanner scr=new FastScanner();
//	static Scanner scr=new Scanner(System.in);
	static PrintStream out=new PrintStream(System.out);
	static StringBuilder sb=new StringBuilder();
	
	
	
	
	static class pair{
		int x;int y;
		pair(int x,int y){ 
			this.x=x;this.y=y;
			
		}
	}
	static class triplet{
		int x;
		long y;
		int z;
		triplet(int x,long y,int z){
			this.x=x;
			this.y=y;
			this.z=z;
		}
	}
	static class FastScanner {
		 BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
	 	String next() {
	 		while (!st.hasMoreTokens())
	 			try {
	 				st=new StringTokenizer(br.readLine());
	 			} catch (IOException e) {
	 				e.printStackTrace();
	 			}
	 			return st.nextToken();
		} long gcd(long a,long b){
			if(b==0) {
				return a;
			}
				
			return gcd(b,a%b);
		}
	 	int[]  sort(int a[]) {
			 
	        int multiplier = 1, len = a.length, max = Integer.MIN_VALUE;
	        int b[] = new int[len];
	        int bucket[];
	        for (int i = 0; i < len; i++) if (max < a[i]) max = a[i];
	        while (max / multiplier > 0) {
	            bucket = new int[10];
	            for (int i = 0; i < len; i++) bucket[(a[i] / multiplier) % 10]++;
	            for (int i = 1; i < 10; i++) bucket[i] += (bucket[i - 1]);
	            for (int i = len - 1; i >= 0; i--) b[--bucket[(a[i] / multiplier) % 10]] = a[i];
	            for (int i = 0; i < len; i++) a[i] = b[i];
	            multiplier *= 10;
	        }
	        return a;
	    }
	 	 long modPow(long base,long exp) {
			if(exp==0) {
				return 1;
			}
			if(exp%2==0) {
				long res=(modPow(base,exp/2));
				return (res*res);
			}
			return (base*modPow(base,exp-1));
		}
		int []reverse(int a[]){
			int b[]=new int[a.length];
			int index=0;
			for(int i=a.length-1;i>=0;i--) {
				b[index]=a[i];
			}
			return b;
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long[] readLongArray(int n) {
			long  [] a=new long  [n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	static void print(Object ele) {
		sb.append(ele).append(" ");
	}
	static void println(Object ele) {
		sb.append(ele).append('\n');
		
	}
	static void println() {
		sb.append('\n');
	}
//	static long LCM(long a,long b) {
//		return a*b/(scr.gcd(a, b));
//	}
//	static long n, m, k, l, q, count, sum, temp, ans;
//	static long []a;
//	static long b[];
//	static  char c[];
//	static char t[];
//	static boolean visited[][];
	
	static int count[];
	static void solve() {
		
		int n = scr.nextInt();
		int k=scr.nextInt();
		int []a=new int[n];
		for(int i=0;i<n;i++) {
			int ele=scr.nextInt();
			a[i]=ele;
		}
		
		PriorityQueue<pair>pq=new PriorityQueue<>((n1,n2)->(n2.x-n1.x));
		long sum=0;
		long cost=0;
		for(int i=0;i<k;i++) {
			pq.add(new pair(a[i],i));
			sum+=a[i];
			cost +=(long)(k-i)*a[i];
		}
		
		int time[]=new int[n];
		
		for(int i=k;i<k+n;i++) {
			if(i<n) {
				pq.add(new pair(a[i],i));
				sum+=a[i];
			}
			
			pair p=pq.poll();
			time[p.y]=i;
			
			sum-=p.x;
			cost+=sum;
		}
		
		sb.append(cost);
		sb.append('\n');
		for(int i:time) {
			sb.append(i+1).append(" ");
		}
	}  
	static int  MAX = Integer.MAX_VALUE;
	static int  MIN = Integer.MIN_VALUE;
	public static void main(String []args) {
//		int t=scr.nextInt();
//		while(t-->0) {
//			solve();
//		}
		solve();
		out.println(sb);
	}
}
 