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.util.*;
import java.io.*;
import java.text.*;
public class Main{
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    void solve(int TC)throws Exception{
        int n = ni();
        Player[] a = new Player[1+n];
        for(int i = 1; i<= n; i++)a[i] = new Player(i, nl());
        Arrays.sort(a, 1, n+1, (Player p1, Player p2) -> Long.compare(p1.v, p2.v));
        int[][] prev = new int[3][1+n];
        long[][] dp = new long[3][1+n];
        dp[0][0] = 0;
        dp[1][0] = IINF;
        dp[2][0] = IINF;
        for(int i = 1; i<= n; i++){
            for(int len = 0; len < 3; len++)dp[len][i] = IINF;
            for(int len = 3; len <= 5; len++){
                if(i-len < 0)continue;
                long sum = dp[(i-len)%3][i-len]+a[i].v-a[i-len+1].v;
                if(sum < dp[i%3][i]){
                    prev[i%3][i] = len;
                    dp[i%3][i] = sum;
                }
            }
        }
        
        int p = n;
        int cnt = 0;
        while(p > 0){
            cnt++;
            int len = prev[p%3][p];
            for(int i = 0; i< len; i++)a[p-i].pos = cnt;
            p -= len;
        }
        pn(dp[n%3][n]+" "+cnt);
        Arrays.sort(a, 1, n+1, (Player p1, Player p2) -> Integer.compare(p1.i, p2.i));
        for(int i = 1; i<= n; i++)p(a[i].pos+" ");
    }
    class Player{
        int i, pos;
        long v;
        public Player(int I, long V){
            i = I;v = V;pos = 0;
        }
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    long IINF = (long)1e13;
    final int INF = (int)1e9, MX = (int)2e6+5;
    DecimalFormat df = new DecimalFormat("0.00");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-7;
    static boolean multipleTC = false, memory = false, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        if(fileIO){
            in = new FastReader("C:/users/user/desktop/inp.in");
            out = new PrintWriter("C:/users/user/desktop/out.out");
        }else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC)?ni():1;
        pre();
        for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}
 
    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }
 
        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }
 
        String nextLine() throws Exception{
            String str = "";
            try{   
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }  
            return str;
        }
    }   
}  