Problem: It's time polar bears Menshykov and Uslada from the zoo of St. Petersburg and elephant Horace from the zoo of Kiev got down to business. In total, there are n tasks for the day and each animal should do each of these tasks. For each task, they have evaluated its difficulty. Also animals decided to do the tasks in order of their difficulty. Unfortunately, some tasks can have the same difficulty, so the order in which one can perform the tasks may vary.

Menshykov, Uslada and Horace ask you to deal with this nuisance and come up with individual plans for each of them. The plan is a sequence describing the order in which an animal should do all the n tasks. Besides, each of them wants to have its own unique plan. Therefore three plans must form three different sequences. You are to find the required plans, or otherwise deliver the sad news to them by stating that it is impossible to come up with three distinct plans for the given tasks.

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of tasks. The second line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 2000), where hi is the difficulty of the i-th task. The larger number hi is, the more difficult the i-th task is.

Output

In the first line print "YES" (without the quotes), if it is possible to come up with three distinct plans of doing the tasks. Otherwise print in the first line "NO" (without the quotes). If three desired plans do exist, print in the second line n distinct integers that represent the numbers of the tasks in the order they are done according to the first plan. In the third and fourth line print two remaining plans in the same form.

If there are multiple possible answers, you can print any of them.

Examples

Input

4
1 3 3 1


Output

YES
1 4 2 3 
4 1 2 3 
4 1 3 2 


Input

5
2 4 1 4 8


Output

NO

Note

In the first sample the difficulty of the tasks sets one limit: tasks 1 and 4 must be done before tasks 2 and 3. That gives the total of four possible sequences of doing tasks : [1, 4, 2, 3], [4, 1, 2, 3], [1, 4, 3, 2], [4, 1, 3, 2]. You can print any three of them in the answer.

In the second sample there are only two sequences of tasks that meet the conditions — [3, 1, 2, 4, 5] and [3, 1, 4, 2, 5]. Consequently, it is impossible to make three distinct sequences of tasks.

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.BufferedWriter;
import java.util.InputMismatchException;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.Writer;
import java.math.BigInteger;
import java.io.InputStream;
import java.util.Arrays;

public class SolB {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        TaskSolver solver = new TaskSolver();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskSolver {

    public void solve(int testNumber, InputReader in, OutputWriter out) {
        int n = in.nextInt();
        Task[] tasks = new Task[n];
        int i = 0;
        for (i = 0; i < n; ++i) {
            tasks[i] = new Task(in.nextInt(), i);
        }
        Arrays.sort(tasks);

        int cnt = 0;
        for (i = 1; i < n; ++i) {
            if(tasks[i].diff == tasks[i-1].diff)
                cnt++;
        }
        if(cnt >= 2)
            System.out.println("YES");
        else{
            System.out.println("NO");
            return;
        }

        for (i = 0; i < n; ++i) {
            System.out.print((tasks[i].pos+1) + " ");
        }
            System.out.println();

        for (i = 1; i < n; ++i) {
            if(tasks[i].diff == tasks[i-1].diff){
                Task temp = tasks[i];
                tasks[i] = tasks[i-1];
                tasks[i-1] = temp;
                break;
            }
        }

        for (i = 0; i < n; ++i) {
            System.out.print((tasks[i].pos+1) + " ");
        }
            System.out.println();

        boolean flag = false;
        for (i = 1; i < n; ++i) {
            if(flag && tasks[i].diff == tasks[i-1].diff){
                Task temp = tasks[i];
                tasks[i] = tasks[i-1];
                tasks[i-1] = temp;
                break;
            }
            if(tasks[i].diff == tasks[i-1].diff){
                flag = true;
            }
        }

        for (i = 0; i < n; ++i) {
            System.out.print((tasks[i].pos+1) + " ");
        }
            System.out.println();

    }
}

/*
    int in.nextInt()
    String in.nextString()

    out.print(n + " ") //must be " " which is a String object.
    out.close()
*/
class InputReader {
    InputStream stream;
    byte[] buf = new byte[1024];
    int curChar, numChars;

    public InputReader(InputStream stream) {
        this.stream = stream;
    }

    private int read() {

        if (numChars == -1)
            throw new InputMismatchException();
        //buffer wraps around here. Hence the buf array size does not matter.
        /*
        if the buffer size is enough, i.e 1024. InputStream load at most 1024 bytes(if necessary)
        into buffer. read() returns 1 byte every time. After the first read, it only needs to
        return corresponding byte in the buf every time it is called. it does not have to call
        input stream to read again.
        Only if the bug has run out does InputReader has to call stream.read(buf) to reload the
        buf.
        */
        if (curChar >= numChars) {
            curChar = 0;
            try {
                numChars = stream.read(buf);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
            if (numChars <= 0)
                return -1;
        }
        return buf[curChar++];
    }


    public int nextInt() {
        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 String nextString() {
        int c = read();
        while (isSpaceChar(c))
            c = read();
        StringBuffer res = new StringBuffer();
        do {
            res.appendCodePoint(c);
            c = read();
        } while (!isSpaceChar(c));
        return res.toString();
    }

    private boolean isSpaceChar(int c) {
        return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
    }

}

class OutputWriter {

    private final PrintWriter writer;

    public OutputWriter(OutputStream outputStream) {
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
    }

    public OutputWriter(Writer writer) {
        this.writer = new PrintWriter(writer);
    }

    public void print(Object... objects) {
        for (int i = 0; i < objects.length; i++) {
            if (i != 0)
                writer.print(' ');
            writer.print(objects[i]);
        }
    }

    public void println(Object... objects){
        print(objects);
        print("\n");
    }

    public void close() {
        writer.close();
    }

}

class Task implements Comparable<Task>{
    public int diff;
    public int pos;
    public Task(int diff, int pos){
        this.diff = diff;
        this.pos = pos;
    }

    public int compareTo(Task t){
        return this.diff - t.diff;
    }
}