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.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.Math.floor;
import static java.lang.Math.ceil;
import static java.lang.Math.pow;
import static java.lang.Math.log;
import static java.lang.Math.sqrt;

/* ------------ Theorems and Lemmas ------------ */

// Fermat's Little Theorem
// (1) ((a ** p) - a) % p = 0; if 'p' is prime
// (2) Further, if 'a % p != 0' (a and b should be coprime), then ((a ** (p - 1)) - 1) % p = 0

/* -------------- End of theorems -------------- */

public class Solution {

	// Digit DP
	private static int digit;
	private static int times;
	private static StringBuilder number;
	private static int[][][] digitDP;
	// End

	protected static final String[] FACTS = new String[] { "", "", "2", "3", "322", "5", "53", "7", "7222", "7332" };

	public static void main(String[] args) {
		FastReader reader = new FastReader();
		// StringBuilder sb = new StringBuilder("");
		int n = reader.nextInt();
		Map<Integer, List<Integer>> map = new HashMap<>();
		for (int i = 1; i <= n; ++i) {
			int val = reader.nextInt();
			if (map.containsKey(val))
				map.get(val).add(i);
			else
				map.put(val, new ArrayList<>(Arrays.asList(i)));
		}
		reader.close();
		// System.out.println(sb.toString().trim());
		List<Integer> list = new ArrayList<>(map.keySet());
		Collections.sort(list);
		StringBuilder res = new StringBuilder("");
		StringBuilder sb = new StringBuilder("");
		Set<String> set = new HashSet<>(3);
		for (int num : list)
			for (int ind : map.get(num))
				sb.append(ind + " ");
		set.add(sb.toString().trim());
		res.append(sb.toString().trim() + "\n");
		sb.setLength(0);
		for (int num : list)
			for (int i = map.get(num).size() - 1; i >= 0; --i)
				sb.append(map.get(num).get(i) + " ");
		set.add(sb.toString().trim());
		res.append(sb.toString().trim() + "\n");
		sb.setLength(0);
		boolean flag = false;
		for (int num : list) {
			if (map.get(num).size() == 1)
				sb.append(map.get(num).get(0) + " ");
			else {
				if (!flag) {
					for (int i = 1, c = 0; c < map.get(num).size(); i = (i + 1) % map.get(num).size(), ++c)
						sb.append(map.get(num).get(i) + " ");
					flag = true;
				} else {
					for (int ind : map.get(num))
						sb.append(ind + " ");
				}
			}
		}
		res.append(sb.toString().trim());
		set.add(sb.toString().trim());
		System.out.println(set.size() == 3 ? "YES\n" + res : "NO");
	}

	public static boolean prime(int n) {
		if (n <= 1)
			return false;
		if (n == 2)
			return true;
		if ((n & 1) == 0)
			return false;
		int val = (int) sqrt((double) n);
		for (int i = 3; i <= val; ++i)
			if (n % i == 0)
				return false;
		return true;
	}

	public static int ddp(int n) {
		number = new StringBuilder(String.valueOf(n));
		digitDP = new int[12][12][2];
		for (int[][] a : digitDP)
			for (int[] b : a)
				Arrays.fill(b, -1);
		return call(0, 0, 0);
	}

	// Digit DP
	public static int call(int pos, int cnt, int f) {
		if (cnt > times)
			return 0;
		if (pos == number.length())
			return cnt == times ? 1 : 0;
		if (digitDP[pos][cnt][f] != -1)
			return digitDP[pos][cnt][f];
		int res = 0;
		int lmt = f == 0 ? number.charAt(pos) - '0' : 9;
		for (int dgt = 0; dgt <= lmt; ++dgt)
			if ((dgt == digit ? 1 + cnt : cnt) <= times)
				res += call(1 + pos, dgt == digit ? 1 + cnt : cnt, f == 0 && dgt < lmt ? 1 : f);
		digitDP[pos][cnt][f] = res;
		return res;
	}

	// Binary Exponentiation
	public static int binpow(int a, int n, int mod) {
		// Iterative approach
		a %= mod;
		int result = 1;
		while (n > 0) {
			if ((n & 1) != 0)
				result = result * a % mod;
			a = a * a % mod;
			n >>= 1;
		}
		return result;

		// Recursive approach
		/*
		 * if (n == 0l) return 1l; long res = binpow(a, n / 2, mod); if ((n & 1l) != 0l)
		 * return ((res % mod) * (res % mod) * (a % mod)) % mod; else return ((res %
		 * mod) * (res % mod)) % mod;
		 */
	}

	// GCD
	public static int gcd(int a, int b) {
		while (b > 0) {
			a %= b;
			int t = a;
			a = b;
			b = t;
		}
		return a;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine().trim());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		float nextFloat() {
			return Float.parseFloat(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		int[] readArray(int size) {
			int[] ar = new int[size];
			for (int i = 0; i < size; ++i)
				ar[i] = nextInt();
			return ar;
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
	}
}