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.BufferedOutputStream;
import java.io.Closeable;
import java.io.DataInputStream;
import java.io.Flushable;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.util.Arrays;
import java.util.Comparator;
import java.util.InputMismatchException;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	static class TaskAdapter implements Runnable {
		@Override
		public void run() {
			long startTime = System.currentTimeMillis();
			InputStream inputStream = System.in;
			OutputStream outputStream = System.out;
			FastReader in = new FastReader(inputStream);
			Output out = new Output(outputStream);
			EYetAnotherDivisionIntoTeams solver = new EYetAnotherDivisionIntoTeams();
			solver.solve(1, in, out);
			out.close();
			System.err.println(System.currentTimeMillis()-startTime+"ms");
		}
	}

	public static void main(String[] args) throws Exception {
		Thread thread = new Thread(null, new TaskAdapter(), "", 1<<28);
		thread.start();
		thread.join();
	}

	static class EYetAnotherDivisionIntoTeams {
		private final int iinf = 1_000_000_000;

		public EYetAnotherDivisionIntoTeams() {
		}

		public void solve(int kase, InputReader in, Output pw) {
			int n = in.nextInt();
			int[][] arr = new int[n][2];
			for(int i = 0; i<n; i++) {
				arr[i][0] = in.nextInt();
				arr[i][1] = i;
			}
			Arrays.sort(arr, Comparator.comparingInt(o -> o[0]));
			int[] dp = new int[n], next = new int[n];
			dp[n-2] = dp[n-1] = iinf;
			int min = arr[n-1][0], ind = n;
			for(int i = n-3; i>=0; i--) {
				next[i] = ind;
				dp[i] = min-arr[i][0];
				int val = dp[i+2]+arr[i+1][0];
				if(val<min) {
					min = val;
					ind = i+2;
				}
			}
			int[] ans = new int[n];
			int i = 0, end = next[0], cnt = 1;
			while(true) {
				for(int j = i; j<end; j++) {
					ans[arr[j][1]] = cnt;
				}
				i = end;
				if(i==n) {
					break;
				}
				end = next[i];
				cnt++;
			}
			pw.println(dp[0], cnt);
			pw.println(ans);
		}

	}

	static class Output implements Closeable, Flushable {
		public StringBuilder sb;
		public OutputStream os;
		public int BUFFER_SIZE;
		public String lineSeparator;

		public Output(OutputStream os) {
			this(os, 1<<16);
		}

		public Output(OutputStream os, int bs) {
			BUFFER_SIZE = bs;
			sb = new StringBuilder(BUFFER_SIZE);
			this.os = new BufferedOutputStream(os, 1<<17);
			lineSeparator = System.lineSeparator();
		}

		public void print(int i) {
			print(String.valueOf(i));
		}

		public void print(String s) {
			sb.append(s);
			if(sb.length()>BUFFER_SIZE >> 1) {
				flushToBuffer();
			}
		}

		public void println(Object... o) {
			for(int i = 0; i<o.length; i++) {
				if(i!=0) {
					print(" ");
				}
				print(String.valueOf(o[i]));
			}
			println();
		}

		public void println() {
			sb.append(lineSeparator);
		}

		public void println(int[] arr) {
			for(int i = 0; i<arr.length; i++) {
				if(i!=0) {
					print(" ");
				}
				print(arr[i]);
			}
			println();
		}

		private void flushToBuffer() {
			try {
				os.write(sb.toString().getBytes());
			}catch(IOException e) {
				e.printStackTrace();
			}
			sb = new StringBuilder(BUFFER_SIZE);
		}

		public void flush() {
			try {
				flushToBuffer();
				os.flush();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

		public void close() {
			flush();
			try {
				os.close();
			}catch(IOException e) {
				e.printStackTrace();
			}
		}

	}

	static class FastReader implements InputReader {
		final private int BUFFER_SIZE = 1<<16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer;
		private int bytesRead;

		public FastReader(InputStream is) {
			din = new DataInputStream(is);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public int nextInt() {
			int ret = 0;
			byte c = skipToDigit();
			boolean neg = (c=='-');
			if(neg) {
				c = read();
			}
			do {
				ret = ret*10+c-'0';
			} while((c = read())>='0'&&c<='9');
			if(neg) {
				return -ret;
			}
			return ret;
		}

		private boolean isDigit(byte b) {
			return b>='0'&&b<='9';
		}

		private byte skipToDigit() {
			byte ret;
			while(!isDigit(ret = read())&&ret!='-') ;
			return ret;
		}

		private void fillBuffer() {
			try {
				bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			}catch(IOException e) {
				e.printStackTrace();
				throw new InputMismatchException();
			}
			if(bytesRead==-1) {
				buffer[0] = -1;
			}
		}

		private byte read() {
			if(bytesRead==-1) {
				throw new InputMismatchException();
			}else if(bufferPointer==bytesRead) {
				fillBuffer();
			}
			return buffer[bufferPointer++];
		}

	}

	static interface InputReader {
		int nextInt();

	}
}

