Polycarp is practicing his problem solving skill. He has a list of n problems with difficulties a_1, a_2, ..., a_n, respectively. His plan is to practice for exactly k days. Each day he has to solve at least one problem from his list. Polycarp solves the problems in the order they are given in his list, he cannot skip any problem from his list. He has to solve all n problems in exactly k days.

Thus, each day Polycarp solves a contiguous sequence of (consecutive) problems from the start of the list. He can't skip problems or solve them multiple times. As a result, in k days he will solve all the n problems.

The profit of the j-th day of Polycarp's practice is the maximum among all the difficulties of problems Polycarp solves during the j-th day (i.e. if he solves problems with indices from l to r during a day, then the profit of the day is max_{l ≤ i ≤ r}a_i). The total profit of his practice is the sum of the profits over all k days of his practice.

You want to help Polycarp to get the maximum possible total profit over all valid ways to solve problems. Your task is to distribute all n problems between k days satisfying the conditions above in such a way, that the total profit is maximum.

For example, if n = 8, k = 3 and a = [5, 4, 2, 6, 5, 1, 9, 2], one of the possible distributions with maximum total profit is: [5, 4, 2], [6, 5], [1, 9, 2]. Here the total profit equals 5 + 6 + 9 = 20.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the number of problems and the number of days, respectively.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2000) — difficulties of problems in Polycarp's list, in the order they are placed in the list (i.e. in the order Polycarp will solve them).

Output

In the first line of the output print the maximum possible total profit.

In the second line print exactly k positive integers t_1, t_2, ..., t_k (t_1 + t_2 + ... + t_k must equal n), where t_j means the number of problems Polycarp will solve during the j-th day in order to achieve the maximum possible total profit of his practice.

If there are many possible answers, you may print any of them.

Examples

Input

8 3
5 4 2 6 5 1 9 2


Output

20
3 2 3

Input

5 1
1 1 1 1 1


Output

1
5


Input

4 2
1 2000 2000 2


Output

4000
2 2

Note

The first example is described in the problem statement.

In the second example there is only one possible distribution.

In the third example the best answer is to distribute problems in the following way: [1, 2000], [2000, 2]. The total profit of this distribution is 2000 + 2000 = 4000.

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;


 
public class TryB
{
    
    static class InputReader {
 
		private final InputStream stream;
		private final byte[] buf = new byte[8192];
		private int curChar, snumChars;
 
		public InputReader(InputStream st) {
			this.stream = st;
		}
 
		public int read() {
			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 ni() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			int res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public long nl() {
			int c = read();
			while (isSpaceChar(c)) {
				c = read();
			}
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = read();
			}
			long res = 0;
			do {
				res *= 10;
				res += c - '0';
				c = read();
			} while (!isSpaceChar(c));
			return res * sgn;
		}
 
		public int[] nia(int n) {
			int a[] = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = ni();
			}
			return a;
		}
 
		public String rs() {
			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 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 boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
 
		private boolean isEndOfLine(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
 
	}
	

	
	
	
    static long mod=1000000007;
    static BigInteger bigInteger = new BigInteger("1000000007");
    static int n = (int)1e6;
    static boolean[] prime;
    static ArrayList<Integer> as;
    static void sieve() {
			Arrays.fill(prime	, true);
			prime[0] = prime[1] = false;
			for(int i = 2 ; i * i <= n ; ++i) {
				if(prime[i]) {
					for(int k = i * i; k< n ; k+=i) {
						prime[k] = false;
					}
				}
			}
			
		}
		static PrintWriter w = new PrintWriter(System.out);
 static char [][]sol;
    public static void main(String[] args)
    {
        
        InputReader sc = new InputReader(System.in);
        //PrintWriter w = new PrintWriter(System.out);
    
       /* 
        prime  = new boolean[n + 1];
		sieve();
		prime[1] = false;
		*/
		/*
		as = new ArrayList<>();
		for(int i=2;i<=1000000;i++)
		{
		    if(prime[i])
		    as.add(i);
		}
		*/
		
       /* 
		
	long a = sc.nl();
	 
            BigInteger ans = new BigInteger("1");
            for (long i = 1; i < Math.sqrt(a); i++) {
                if (a % i == 0) {
                    if (a / i == i) {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        
                    } else {
                        ans = ans.multiply(BigInteger.valueOf(phi(i)));
                        ans = ans.multiply(BigInteger.valueOf(phi(a / i)));
 
                        
                    }
 
 
                }
            }
            w.println(ans.mod(bigInteger));
            */
          
         // MergeSort ob = new MergeSort();
       // ob.sort(arr, 0, arr.length-1);
       
    // Student []st = new Student[x];    
   // st[i] = new Student(i,d[i]);
     //Arrays.sort(st,(p,q)->p.diff-q.diff);
     
     
     int x = sc.ni();
     int k = sc.ni();
     int []a = sc.nia(x);
     ArrayList<Integer> b = new ArrayList<>();
     ArrayList<Integer> b1 = new ArrayList<>();
    for(int i=0;i<x;i++)
    {
        b.add(a[i]);
    }
     Arrays.sort(a);
     long sum = 0;
     int p = x-1;
     while(k-- > 0)
     {
         sum += (long) a[p];
         b1.add(a[p]);
         p--;
         
     }
     w.println(sum);
     ArrayList<Integer> b2 = new ArrayList<>();
     int []vis = new int[200001];
    for(int i=0;i<b1.size();i++)
    {
        vis[b1.get(i)]++;
    }
     int curr = 0;
     for(int i=0;i<b.size();i++)
     {
         if(b1.contains(b.get(i)) && vis[b.get(i)] > 0)
         {
             b2.add(i+1);
             vis[b.get(i)]--;
         }
     }
     w.println();
      for(int i=0;i<b2.size()-1;i++)
     {
         w.print((b2.get(i) - curr) + " ");
         curr = b2.get(i);
     }
     w.print(x-curr);
     
     
   
     
     
     

    
    
    
    
    
   w.close();
        
  
    
    }
   
     static class Student
    {
        
        int first;
        int sec;
        Student(int first,int sec)
        {
            this.first = first;
            this.sec = sec;
            
            
        }
        
    }
     
     public static long modMultiply(long one, long two) {
 
		return (one % mod * two % mod) % mod;
	}
    static long fx(int m)
    {
        long re = 0;
        for(int i=1;i<=m;i++)
        {
            re += (long) (i / gcd(i,m));
        }
        return re;
    }
     static long gcd(long a, long b)
    {
	 if (a == 0)
	  return b;
				         
	return gcd(b%a, a);
    }
    
   
    
     static long phi(long nx)
    {
        // Initialize result as n
        double result = nx; 
        
     
        // Consider all prime factors of n and for 
        // every prime factor p, multiply result
        // with (1 - 1/p)
        for (int p = 0; as.get(p) * as.get(p) <= nx; ++p)
        {
            // Check if p is a prime factor.
            if (nx % as.get(p) == 0)
            {
                // If yes, then update n and result
                while (nx % as.get(p) == 0)
                    nx /= as.get(p);
                result *= (1.0 - (1.0 / (double) as.get(p)));
            }
        }
     
        // If n has a prime factor greater than sqrt(n)
        // (There can be at-most one such prime factor)
        if (nx > 1)
            result *= (1.0 - (1.0 / (double) nx));
     
        return (long)result;
    
        
        //return(phi((long)result,k-1));
        
    }
    
    public static void primeFactors(int n,int x)
    {
        as = new ArrayList<>();
       
        // Print the number of 2s that divide n
        while (n%2==0)
        {
            
        
            as.add(2);
            //System.out.print(2 + " ");
            n /= 2;
            
        }
 
        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i <= Math.sqrt(n); i+= 2)
        {
            // While i divides n, print i and divide n
            while (n%i == 0)
            {
            
               // System.out.print(i + " ");
                
                as.add(i);
                n /= i;
             
                
            }
            
        }
 
        // This condition is to handle the case whien
        // n is a prime number greater than 2
        if (n >= 2)
        {
          
            as.add(n);
        }
           
    }
    static int digitsum(int x)
    
    {    
 
        int sum = 0;
        while(x > 0)
        {
            int temp = x % 10;
            sum += temp;
            x /= 10;
        }
        return sum;
        
    }
    static int countDivisors(int n)
    {
        int cnt = 0;
        for (int i = 1; i*i <=n; i++)
        {
             if (n % i == 0 && i<=1000000)
             {
                // If divisors are equal,
                // count only one
                if (n / i == i)
                   cnt++;
      
                else // Otherwise count both
                    cnt = cnt + 2;
             }
         }
        return cnt;
    }
   
    static boolean isprime(int n)
    {
    
    if(n == 2)
        return true;
    if(n == 3)
        return true;
    if(n % 2 == 0)
        return false;
    if(n % 3 == 0)
        return false;
 
    int i = 5;
   int w = 2;
 
    while(i * i <= n)
    {
        if(n % i == 0)
            return false;
    
 
        i += w;
        w = 6 - w;
    }
 
    return true;
    }
    
    static long log2(long value) {
    return Long.SIZE-Long.numberOfLeadingZeros(value);
}
 
 
 
 
    
    
    static boolean binarysearch(int []arr,int p,int n)
    {
        //ArrayList<Integer> as = new ArrayList<>();
        //as.addAll(0,at);
        //Collections.sort(as);
        boolean re = false;
        int st = 0;
        int end = n-1;
        
        
        
        while(st <= end)
        { 
            int mid = st + (end-st)/2;
        
            if(p > arr[mid])
            {
                st = mid+1;
                
            }
            else if(p < arr[mid])
            {
                end = mid-1;
                
            }
            else if(p == arr[mid])
            {
            
            re = true;
            break;
            }
           
            
            
        }
         return re;
        
    }
    
   
    
    /* Java program for Merge Sort */
static class MergeSort
{
    // Merges two subarrays of arr[].
    // First subarray is arr[l..m]
    // Second subarray is arr[m+1..r]
    void merge(int arr[], int l, int m, int r)
    {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;
 
        /* Create temp arrays */
        int L[] = new int [n1];
        int R[] = new int [n2];
 
        /*Copy data to temp arrays*/
        for (int i=0; i<n1; ++i)
            L[i] = arr[l + i];
        for (int j=0; j<n2; ++j)
            R[j] = arr[m + 1+ j];
 
 
        /* Merge the temp arrays */
 
        // Initial indexes of first and second subarrays
        int i = 0, j = 0;
 
        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2)
        {
            if (L[i] <= R[j])
            {
                arr[k] = L[i];
                i++;
            }
            else
            {
                arr[k] = R[j];
                j++;
            }
            k++;
        }
 
        /* Copy remaining elements of L[] if any */
        while (i < n1)
        {
            arr[k] = L[i];
            i++;
            k++;
        }
 
        /* Copy remaining elements of R[] if any */
        while (j < n2)
        {
            arr[k] = R[j];
            j++;
            k++;
        }
    }
 
    // Main function that sorts arr[l..r] using
    // merge()
    void sort(int arr[], int l, int r)
    {
        if (l < r)
        {
            // Find the middle point
            int m = (l+r)/2;
 
            // Sort first and second halves
            sort(arr, l, m);
            sort(arr , m+1, r);
 
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }
 
    /* A utility function to print array of size n */
    
 
   
}
    

    
    
    
    public static int ip(String s){
		return Integer.parseInt(s);
	}
	 public static String ips(int s){
		return Integer.toString(s);
	}
	
    
					 
}