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.*;
import java.math.*;
public class Main
{
    static class Reader 
    { 
        private InputStream mIs;private byte[] buf = new byte[1024];private int curChar,numChars;public Reader() { this(System.in); }public Reader(InputStream is) { mIs = is;} 
        public int read() {if (numChars == -1) throw new InputMismatchException();if (curChar >= numChars) {curChar = 0;try { numChars = mIs.read(buf);} catch (IOException e) { throw new InputMismatchException();}if (numChars <= 0) return -1; }return buf[curChar++];} 
        public String nextLine(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isEndOfLine(c));return res.toString() ;} 
        public String s(){int c = read();while (isSpaceChar(c)) c = read();StringBuilder res = new StringBuilder();do {res.appendCodePoint(c);c = read();}while (!isSpaceChar(c));return res.toString();} 
        public long l(){int c = read();while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }long res = 0; do{ if (c < '0' || c > '9') throw new InputMismatchException();res *= 10 ; res += c - '0' ; c = read();}while(!isSpaceChar(c));return res * sgn;} 
        public int i(){int c = read() ;while (isSpaceChar(c)) c = read();int sgn = 1;if (c == '-') { sgn = -1 ; c = read() ; }int res = 0;do{if (c < '0' || c > '9') throw new InputMismatchException();res *= 10 ; res += c - '0' ; c = read() ;}while(!isSpaceChar(c));return res * sgn;} 
        public double d() throws IOException {return Double.parseDouble(s()) ;}
        public boolean isSpaceChar(int c) { return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1; } 
        public boolean isEndOfLine(int c) { return c == '\n' || c == '\r' || c == -1; } 
        public int[] arr(int n){int[] ret = new int[n];for (int i = 0; i < n; i++) {ret[i] = i();}return ret;}
    }
    
    
    ///////////////////////////////////////////////////////////////////////////////////////////
    //    RRRRRRRRR            AAA             HHH     HHH     IIIIIIIIIIIII   LLL           //
    //    RR    RRR           AAAAA            HHH     HHH      IIIIIIIIIII    LLL           //
    //    RR   RRR           AAAAAAA           HHH     HHH          III        LLL           //
    //    RR  RRR           AAA   AAA          HHHHHHHHHHH          III        LLL           //
    //    RRRRRR           AAA     AAA         HHHHHHHHHHH          III        LLL           //
    //    RR  RRR         AAAAAAAAAAAAA        HHH     HHH          III        LLL           //
    //    RR    RRR      AAA         AAA       HHH     HHH      IIIIIIIIIII    LLLLLLLLLLLL  //
    //    RR     RRR    AAA           AAA      HHH     HHH     IIIIIIIIIIIII   LLLLLLLLLLLL  //
    ///////////////////////////////////////////////////////////////////////////////////////////
    static class pair 
    {
       int x;
       int y;
       public pair (int k, int p) 
       {
           x = k;
           y = p;
       }
       @Override
       public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            pair pair = (pair) o;
            return x == pair.x && y == pair.y;
        }
    
       @Override
       public int hashCode() {
            return Objects.hash(x, y);
       }
    }
    static class PriorityComparator implements Comparator<pair>
    {
        public int compare(pair first, pair second) 
        {
            return second.x-first.x;
        }
    }
    public static void main(String[] args)throws IOException
    {
        PrintWriter out= new PrintWriter(System.out);
        Reader sc=new Reader();
        int n=sc.i(),k=sc.i(),arr[]=sc.arr(n);
        PriorityComparator pc = new PriorityComparator();
        PriorityQueue<pair> pq = new PriorityQueue<pair>(pc);
        TreeSet<Integer> ts=new TreeSet<>();
        for(int i=1+k;i<=n+k;i++)ts.add(i);
        for(int i=0;i<n;i++)pq.add(new pair(arr[i],i));
        int ans[]=new int[n];
        long sum=0;
        while(pq.size()!=0)
        {
            pair p=pq.remove();
            sum+=(long)arr[p.y]*(ts.higher(p.y)-(p.y+1));
            ans[p.y]=ts.higher(p.y);
            ts.remove(ts.higher(p.y));
        }
        out.println(sum);
        for(int i=0;i<n;i++)out.print(ans[i]+" ");
        out.flush();
    }
}