Problem: Don't you tell me what you think that I can be

If you say that Arkady is a bit old-fashioned playing checkers, you won't be right. There is also a modern computer game Arkady and his friends are keen on. We won't discuss its rules, the only feature important to this problem is that each player has to pick a distinct hero in the beginning of the game.

There are 2 teams each having n players and 2n heroes to distribute between the teams. The teams take turns picking heroes: at first, the first team chooses a hero in its team, after that the second team chooses a hero and so on. Note that after a hero is chosen it becomes unavailable to both teams.

The friends estimate the power of the i-th of the heroes as p_i. Each team wants to maximize the total power of its heroes. However, there is one exception: there are m pairs of heroes that are especially strong against each other, so when any team chooses a hero from such a pair, the other team must choose the other one on its turn. Each hero is in at most one such pair.

This is an interactive problem. You are to write a program that will optimally choose the heroes for one team, while the jury's program will play for the other team. Note that the jury's program may behave inefficiently, in this case you have to take the opportunity and still maximize the total power of your team. Formally, if you ever have chance to reach the total power of q or greater regardless of jury's program choices, you must get q or greater to pass a test.

Input

The first line contains two integers n and m (1 ≤ n ≤ 10^3, 0 ≤ m ≤ n) — the number of players in one team and the number of special pairs of heroes.

The second line contains 2n integers p_1, p_2, …, p_{2n} (1 ≤ p_i ≤ 10^3) — the powers of the heroes.

Each of the next m lines contains two integer a and b (1 ≤ a, b ≤ 2n, a ≠ b) — a pair of heroes that are especially strong against each other. It is guaranteed that each hero appears at most once in this list.

The next line contains a single integer t (1 ≤ t ≤ 2) — the team you are to play for. If t = 1, the first turn is yours, otherwise you have the second turn.

Hacks

In order to hack, use the format described above with one additional line. In this line output 2n distinct integers from 1 to 2n — the priority order for the jury's team. The jury's team will on each turn select the first possible hero from this list. Here possible means that it is not yet taken and does not contradict the rules about special pair of heroes.

Interaction

When it is your turn, print a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by you. Note that you can't choose a hero previously chosen by either you of the other player, and you must follow the rules about special pairs of heroes.

When it is the other team's turn, read a line containing a single integer x (1 ≤ x ≤ 2n) — the index of the hero chosen by the other team. It is guaranteed that this index is not chosen before and that the other team also follows the rules about special pairs of heroes.

After the last turn you should terminate without printing anything.

After printing your choice do not forget to output end of line and flush the output. Otherwise you will get Idleness limit exceeded. To do this, use:

  * fflush(stdout) or cout.flush() in C++;
  * System.out.flush() in Java;
  * flush(output) in Pascal;
  * stdout.flush() in Python;
  * see documentation for other languages.



Jury's answer -1 instead of a valid choice means that you made an invalid turn. Exit immediately after receiving -1 and you will see Wrong answer verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream.

Examples

Input


3 1
1 2 3 4 5 6
2 6
1

2

4

1


Output






6

5

3


Input


3 1
1 2 3 4 5 6
1 5
2
6

1

3


Output







5

4

2

Note

In the first example the first turn is yours. In example, you choose 6, the other team is forced to reply with 2. You choose 5, the other team chooses 4. Finally, you choose 3 and the other team choose 1.

In the second example you have the second turn. The other team chooses 6, you choose 5, forcing the other team to choose 1. Now you choose 4, the other team chooses 3 and you choose 2.

Solution (in JAVA):

Solution:
import static java.util.Arrays.*;
import java.util.stream.*;

public class C {

	public Object solve() {
		int N = sc.nextInt(), M = sc.nextInt();
		int [] P = sc.nextInts();
		int [][] Z  = dec(sc.nextInts(M));

		int [] C = new int [2*N]; fill(C, -1);
		for (int [] z : Z)
			for (int j : rep(2))
				C[z[j]] = z[1-j];

		int T = sc.nextInt();
		boolean [] V = new boolean [2*N];

		int x = -1, SZ = N;
		if (T == 2) {
			x = write(-1, V);
			SZ = N-1;
		}

		next: for (int i : rep(SZ)) {
			if (x >= 0 && C[x] >= 0 && !V[C[x]])
				x = write(C[x], V);
			else {
				for (int j : rep(2*N))
					if (!V[j] && C[j] >= 0 && P[j] >= P[C[j]]) {
						x = write(j, V);
						continue next;
					}
				int best = -1;
				for (int j : rep(2*N))
					if (!V[j] && (best == -1 || P[j] > P[best]))
						best = j;
				x = write(best, V);

			}
		}


		for (int i : rep(2*N))
			if (!V[i]) {
				System.out.println(i+1);
				System.out.flush();
			}

		return null;
	}

	int write (int N, boolean [] V) {
		if (N >= 0) {
			V[N] = true;
			System.out.println(N+1);
			System.out.flush();
		}
		int res = sc.nextInt() - 1;
		V[res] = true;
		return res;
	}

	private static final boolean ONE_TEST_CASE = true;

	private static void init() {
	}

	private static final int INF = (int) 1e9 + 10;
	private static int [][] dec (int [] ... E) { return dec(E, INF); }
	private static int [][] dec (int [][] E, int N) { for (int [] e : E) for (int i = 0; i < e.length && i < N; ++i) --e[i]; return E; }
	private static int [] rep (int N) { return rep(0, N); }
	private static int [] rep (int S, int T) { if (S >= T) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	//////////////////////////////////////////////////////////////////////////////////// OFF
	private static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static class IOUtils {
		public static class MyScanner {
			public String next() { newLine(); return line[index++]; }
			public int nextInt() { return Integer.parseInt(next()); }
			public String nextLine() { line = null; return readLine(); }
			public String [] nextStrings() { return split(nextLine()); }
			public int[] nextInts() { return nextStream().mapToInt(Integer::parseInt).toArray(); }
			public int[][] nextInts(int N) { return IntStream.range(0, N).mapToObj(i -> nextInts()).toArray(int[][]::new); }
			//////////////////////////////////////////////
			private boolean eol() { return index == line.length; }
			private String readLine() {
				try {
					return r.readLine();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private final java.io.BufferedReader r;
			private MyScanner () { this(new java.io.BufferedReader(new java.io.InputStreamReader(System.in))); }
			private MyScanner (java.io.BufferedReader r) {
				try {
					this.r = r;
					while (!r.ready())
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine() {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private java.util.stream.Stream<String> nextStream() { return java.util.Arrays.stream(nextStrings()); }
			private String [] split(String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build(Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim(String delim, Object o, Object ... A) {
			StringBuilder b = new StringBuilder();
			append(b, o, delim);
			for (Object p : A)
				append(b, p, delim);
			return b.substring(delim.length());
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static final java.text.DecimalFormat formatter = new java.text.DecimalFormat("#.#########");
		private static void start () { if (t == 0) t = millis(); }
		private static void append (java.util.function.Consumer<Object> f, java.util.function.Consumer<Object> g, final Object o) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					f.accept(java.lang.reflect.Array.get(o, i));
			}
			else if (o instanceof Iterable<?>)
				((Iterable<?>)o).forEach(f::accept);
			else
				g.accept(o instanceof Double ? formatter.format(o) : o);
		}
		private static void append (final StringBuilder b, Object o, final String delim) {
			append(x -> { append(b, x, delim); }, x -> b.append(delim).append(x), o);
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void err (Object o, Object ... A) { System.err.println(build(o, A)); }
		private static boolean DEBUG, PRINT;
		private static void write (Object o) {
			err(o, '(', time(), ')');
			if (PRINT)
				pw.println(o);
		}
		private static void exit () {
			IOUtils.pw.close();
			System.out.flush();
			err("------------------");
			err(time());
			System.exit(0);
		}
		private static long t;
		private static long millis() { return System.currentTimeMillis(); }
		private static String time() { return "Time: " + (millis() - t) / 1000.0; }
		private static void run(int N) {
			try {
				DEBUG = System.getProperties().containsKey("DEBUG");
				PRINT = System.getProperties().containsKey("PRINT");
			}
			catch (Throwable t) {}

			for (int n = 1; n <= N; ++n) {
				Object res = new C().solve();
				if (res != null)
					write("Case #" + n + ": " + build(res));
			}
			exit();
		}
	}
	////////////////////////////////////////////////////////////////////////////////////
	public static void main(String[] args) {
		init();
		int N = ONE_TEST_CASE ? 1 : sc.nextInt();
		IOUtils.run(N);
	}
}