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.*;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.List;

public class Main {
    private final Reader reader;

    private final Writer writer;

    public static void main(String[] args) throws IOException {
        new Main().solve();

    }

    private void solve() throws IOException {
        int N = reader.nextUnsignedInt();
        Pair pairs[] = new Pair[N];

        for (int i = 0; i < N; i++) {
            pairs[i] = new Pair(reader.nextUnsignedInt(), i + 1);
        }

        Arrays.sort(pairs);

        List<Integer> swaps = new LinkedList<Integer>();

        for (int i = 1; i < N && swaps.size() < 2; i++) {
            if (pairs[i].difficulty == pairs[i - 1].difficulty) {
                swaps.add(i);
            }
        }

        if (swaps.size() < 2) {
            writer.print("NO\n");
        writer.flush();
            return;
        }

        writer.print("YES\n");
        print(pairs);
        swap(pairs, swaps.get(0), swaps.get(0) - 1);
        print(pairs);
        swap(pairs, swaps.get(1), swaps.get(1) - 1);
        print(pairs);
        writer.flush();
    }

    private void print(Pair[] pairs) throws IOException {
        for (int i = 0; i < pairs.length - 1; i++) {
            writer.print(pairs[i].index + " ");
        }
        writer.print(pairs[pairs.length - 1].index + "\n");
    }

    private void swap(Pair[] pairs, int i1, int i2) {
        Pair tmp = pairs[i1];
        pairs[i1] = pairs[i2];
        pairs[i2] = tmp;
    }

    private static class Pair implements Comparable<Pair> {
        int difficulty;
        int index;

        public Pair(int difficulty, int index) {
            this.difficulty = difficulty;
            this.index = index;
        }

        @Override
        public int compareTo(Pair o) {
            return this.difficulty - o.difficulty;
        }
    }

    Main(InputStream in, OutputStream out) {
        this.reader = new Reader(in);
        this.writer = new Writer(out);
    }

    public Main() {
        this.reader = new Reader();
        this.writer = new Writer();
    }


}

class Reader {

    private final int BUFFER_SIZE = 65536;

    private final DataInputStream din;

    private final byte[] buffer;

    private int bufferPointer, bytesRead;

    Reader() {
        din = new DataInputStream(System.in);
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

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

    public long nextUnsignedLong() throws IOException {
        long ret = 0;
        byte c = read();
        while (c <= 32) c = read();
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0');
        return ret;
    }

    public int nextUnsignedInt() throws IOException {
        int ret = 0;
        byte c = read();
        while (c <= 32) c = read();
        do {
            ret = ret * 10 + c - '0';
        } while ((c = read()) >= '0');
        return ret;
    }

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

    public char firstCharOfNextUnSpacedString() throws IOException {
        byte c;
        char res;
        while ((c = read()) < 33) ;
        res = (char) c;
        while ((c = read()) >= 33) ;
        return res;
    }

    private void fillBuffer() throws IOException {
        bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
        if (bytesRead == -1) {
            buffer[0] = -1;
        }
    }

    private byte read() throws IOException {
        if (bufferPointer == bytesRead) {
            fillBuffer();
        }
        return buffer[bufferPointer++];
    }
}

class Writer {
    private final BufferedWriter out;

    Writer() {
        out = new BufferedWriter(new OutputStreamWriter(System.out), 512);
    }

    public Writer(OutputStream out) {
        this.out = new BufferedWriter(new OutputStreamWriter(out), 512);
    }

    void print(char c) throws IOException {
        out.write(c);
    }

    void print(String str) throws IOException {
        out.write(str);
    }

    void println() throws IOException {
        out.write("\n");
    }

    void flush() throws IOException {
        out.flush();
    }

    @Override
    protected void finalize() throws IOException {
        out.flush();
    }
}