There are n students at your university. The programming skill of the i-th student is a_i. As a coach, you want to divide them into teams to prepare them for the upcoming ICPC finals. Just imagine how good this university is if it has 2 ⋅ 10^5 students ready for the finals!

Each team should consist of at least three students. Each student should belong to exactly one team. The diversity of a team is the difference between the maximum programming skill of some student that belongs to this team and the minimum programming skill of some student that belongs to this team (in other words, if the team consists of k students with programming skills a[i_1], a[i_2], ..., a[i_k], then the diversity of this team is max_{j=1}^{k} a[i_j] - min_{j=1}^{k} a[i_j]).

The total diversity is the sum of diversities of all teams formed.

Your task is to minimize the total diversity of the division of students and find the optimal way to divide the students.

Input

The first line of the input contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the number of students.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the programming skill of the i-th student.

Output

In the first line print two integers res and k — the minimum total diversity of the division of students and the number of teams in your division, correspondingly.

In the second line print n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ k), where t_i is the number of team to which the i-th student belong.

If there are multiple answers, you can print any. Note that you don't need to minimize the number of teams. Each team should consist of at least three students.

Examples

Input


5
1 1 3 4 2


Output


3 1
1 1 1 1 1 


Input


6
1 5 12 13 2 15


Output


7 2
2 2 1 1 2 1 


Input


10
1 2 5 129 185 581 1041 1909 1580 8150


Output


7486 3
3 3 3 2 2 2 2 1 1 1 

Note

In the first example, there is only one team with skills [1, 1, 2, 3, 4] so the answer is 3. It can be shown that you cannot achieve a better answer.

In the second example, there are two teams with skills [1, 2, 5] and [12, 13, 15] so the answer is 4 + 3 = 7.

In the third example, there are three teams with skills [1, 2, 5], [129, 185, 581, 1041] and [1580, 1909, 8150] so the answer is 4 + 912 + 6570 = 7486.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static  ArrayList<Integer>adj[];
	static PrintWriter out=new PrintWriter(System.out);
	 public static int mod = 1000000007;

	static Boolean notmemo[][][];
	static int k;
	static Pair a[];
	static long b[];
	static int m;
	static int trace[];
	public static void main(String[] args) throws Exception {
	Scanner sc=new Scanner(System.in);
	n=sc.nextInt();
	a=new Pair[n];
	for (int i = 0; i < a.length; i++) {
		a[i]=new Pair(i,sc.nextInt());
	}
	Arrays.sort(a);
	memo=new long[n][8];
	for (int i = 0; i < memo.length; i++) {
		Arrays.fill(memo[i],-1);
	}
	
	out.print(dp(0,0));
	trace=new int[n];
	
	trace(0,0);
	out.print(" "+teams);
	out.println();
	for (int i = 0; i < trace.length; i++) {
		out.print(trace[i]+" ");
	}
	out.flush();
	}
	static int teams=0;
private static void trace(int i, int count) {
		if(i==n) 
			return;
		if(count>7) {
			count=7;
		}

		
		long ans=dp(i,count);
		if(count>=2&&(ans==dp(i+1,0)+a[i].val)) {
		    trace[(int) a[i].idx] = teams;
            trace(i+1, 0);
            return;
		}
		  if (count == 0) teams++;
	        trace[(int) (a[i].idx)] = teams;
	        trace(i+1,count + 1);
	
}
static long mini[];
static long x[];
static long s[];

static long count=0;
	static long dp(int i,int count) {
		if(i==n) {
			if(count!=0) {
				return INF;
			}
			else {
				return 0;
			}
		}
		if(count>7) {
			count=7;
		}
		if(memo[i][count]!=-1) {
			return memo[i][count];
		}
		long ans=INF;
		if(count==0) {
			ans=Math.min(ans, dp(i+1,1)+(-a[i].val));	
		}
		else if(count>0) {
			ans=Math.min(ans, dp(i+1,count+1));		
		}
		
		if(count>=2) {
			ans=Math.min(ans, dp(i+1,0)+a[i].val);
		
		}
		return memo[i][count]=ans;
    }
	static class incpair implements Comparable<incpair>
{
	int a; long b;
	int idx;
	incpair(int a, long dirg,int i) {this.a = a;	b = dirg; idx=i; }
	
	public int compareTo(incpair e){ return (int) (b - e.b);	}
}
	static class decpair implements Comparable<decpair>
{
	int a; long b;int idx;
	
	decpair(int a, long dirg,int i) {this.a = a;	b = dirg; idx=i;}
	
	public int compareTo(decpair e){ return (int) (e.b- b);	}
}

	
	
	
	
	
	


	static long allpowers[];
	static class Quad implements Comparable<Quad>{
		int u;
		int v;
		char state;
		int turns;
		public Quad(int i, int j, char c, int k) {
		u=i;
		v=j;
		state=c;
		turns=k;
		}
		public int compareTo(Quad e){ return (int) (turns - e.turns);	}
		
	}
	static long dirg[][];
	static Edge[] driver;
	
	static int n;
		static class Edge implements Comparable<Edge>
	{
		int node; long cost;
		
		Edge(int a, long dirg) { node = a;	cost = dirg; }
		
		public int compareTo(Edge e){ return (int) (cost - e.cost);	}
	}

	



	  static long manhatandistance(long x,long x2,long y,long y2) {
		  return  Math.abs(x-x2)+Math.abs(y-y2);
	  }
	  static long fib[];
	  static long fib(int n) {
		  if(n==1||n==0) {
			  return 1;
		  }
		  if(fib[n]!=-1) {
			  return fib[n];
		  }
		  else 
			  return  fib[n]=  ((fib(n-2)%mod+fib(n-1)%mod)%mod);
	  }
	  static class Pair implements Comparable<Pair>{
		  long val;
		  int idx;
		  Pair(int a, long b){
			  idx=a;
			  val=b;
		  }
		public int compareTo(Pair p) {
		
			return (int) (this.val-p.val);
		}
		  
		  
	
		
		
	}
	static long[][] comb;		
	static long nCr1(int n , int k)
	{
		if(n < k)
			return 0;
		if(k == 0 || k == n)
			return 1;
		if(comb[n][k] != -1)
			return comb[n][k];
		if(n - k < k)
			return comb[n][k] = nCr1(n, n - k);
		return comb[n][k] = (nCr1(n - 1, k - 1) + nCr1(n - 1, k))%mod;
	}
	
	
static  class Triple implements Comparable<Triple>{

int u;
int v;
		long totalcost;
		public Triple(int a,int b,long l) {
			u=a;
			v=b;
			totalcost=l;
		}
		@Override
		public int compareTo(Triple o) {
			return Long.compare( totalcost,o.totalcost);	
		}
		
	}
	
	static TreeSet<Long> primeFactors(long N)		// O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(p * p <= N)
		{
			while(N % p == 0) { factors.add((long) p); N /= p; }
			if(primes.size()>idx+1)
			p = primes.get(++idx);
			else
				break;
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}
	
	static boolean visited[];
	/**static int bfs(int s)
	{
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		int count=0;
		int maxcost=0;
		int dist[]=new int[n];
		dist[s]=0;
		while(!q.isEmpty())
		{
			
			int u = q.remove();
			if(dist[u]==k) {
				break;
			}
			for(Pair v: adj[u])
			{	
				maxcost=Math.max(maxcost, v.cost);
			
			
			
			if(!visited[v.v])	{
				
				visited[v.v]=true;
				q.add(v.v);
				dist[v.v]=dist[u]+1;
				maxcost=Math.max(maxcost, v.cost);
			}
			}	
			
		}
		return maxcost;
	}**/
	public static boolean FindAllElements(int n, int k) {
	 int sum = k;  
     int[] A = new int[k];  
      Arrays.fill(A, 0, k, 1);  
        
      for (int i = k - 1; i >= 0; --i) {  
      
           while (sum + A[i] <= n) {  
      
              sum += A[i];  
              A[i] *= 2;  
          }  
      }  
      if(sum==n) {
      	return true;
      }
      else
    	return false; 
	}
	
	static long[][] memo;

	
	static boolean vis2[][];
	
	static class SegmentTree {		// 1-based DS, OOP
		
		int N; 			//the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;
		
		SegmentTree(int[] in)		
		{
			array = in; N = in.length - 1;
			sTree = new int[N<<1];		//no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N<<1];
			build(1,1,N);
		}
		
		void build(int node, int b, int e)	// O(n)
		{
			if(b == e)					
				sTree[node] = array[b];
			else						
			{
				int mid = b + e >> 1;
				build(node<<1,b,mid);
				build(node<<1|1,mid+1,e);
				sTree[node] = sTree[node<<1]+sTree[node<<1|1];
			}
		}
		
		
		void update_point(int index, int val)			// O(log n)
		{
			index += N - 1;				
			sTree[index] += val;			
			while(index>1)				
			{
				index >>= 1;
				sTree[index] = sTree[index<<1] + sTree[index<<1|1];		
			}
		}
		
		
		void update_range(int i, int j, int val)		// O(log n) 
		{
			update_range(1,1,N,i,j,val);
		}
		
		void update_range(int node, int b, int e, int i, int j, int val)
		{
			if(i > e || j < b)		
				return;
			if(b >= i && e <= j)		
			{
				sTree[node] += (e-b+1)*val;			
				lazy[node] += val;				
			}							
			else		
			{
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node<<1,b,mid,i,j,val);
				update_range(node<<1|1,mid+1,e,i,j,val);
				sTree[node] = sTree[node<<1] + sTree[node<<1|1];		
			}
			
		}
		void propagate(int node, int b, int mid, int e)		
		{
			lazy[node<<1] += lazy[node];
			lazy[node<<1|1] += lazy[node];
			sTree[node<<1] += (mid-b+1)*lazy[node];		
			sTree[node<<1|1] += (e-mid)*lazy[node];		
			lazy[node] = 0;
		}
		
		int query(int i, int j)
		{
			return query(1,1,N,i,j);
		}
		
		int query(int node, int b, int e, int i, int j)	// O(log n)
		{
			if(i>e || j <b)
				return 0;		
			if(b>= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			int q1 = query(node<<1,b,mid,i,j);
			int q2 = query(node<<1|1,mid+1,e,i,j);
			return q1 + q2;	
					
		}
	
	}
	static boolean f2=false;
	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r)	//C(p x r) = A(p x q) x (q x r) -- O(p x q x r)
	{
		long[][] C = new long[p][r];
		for(int i = 0; i < p; ++i) {
			for(int j = 0; j < r; ++j) {
				for(int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j]+(a2[i][k]%mod * b[k][j]%mod))%mod;
		 C[i][j]%=mod;
				}
		
			}
		}
		return C;
	}
		
	
	
		
		
		
		

		
	
	static ArrayList<Pair> a1[];
	static int memo1[];
	
	static boolean vis[];
	static TreeSet<Integer> set=new TreeSet<Integer>();
	
	static long modPow(long ways, long count, long mod)	// O(log e)
	{
		ways %= mod;
		long res = 1;
		while(count > 0)
		{
			if((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res%mod;
	}
	
	
	

	static long gcd(long ans,long b) {
		if(b==0) {
			return ans;
		}
		return gcd(b,ans%b);
	}
	static int[] isComposite;
	static int[] valid;
	
	static ArrayList<Integer> primes;
	static ArrayList<Integer> l;
	static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
		l=new ArrayList<>();
		valid=new int[N+1];
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				l.add(i);
				valid[i]=1;
				
				for (int j = i*2; j <=N; j +=i) {	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
						
						
						
				      
					}
			}
		
		for(int i=0 ; i<primes.size() ; i++) {
			for(int j:primes) {
				if(primes.get(i)*j>N) {
					break;
				}
				valid[primes.get(i)*j]=1;
			}
		}
	}
	
	public static long[] schuffle(long[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x=(int)(Math.random()*a2.length);
		long temp=a2[x];
		a2[x]=a2[i];
		a2[i]=temp;
		}
		return a2;
	}
		static int V;
	static long INF=(long) 1E18;
		
		static class Edge2 
		{
			int node;
		long cost;
			long next;
			Edge2(int a, int c,Long long1) { node = a;	cost = long1; next=c;}
			
		}
static class Scanner {
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}	
}