Problem: This is an interactive problem!

An arithmetic progression or arithmetic sequence is a sequence of integers such that the subtraction of element with its previous element (x_i - x_{i - 1}, where i ≥ 2) is constant — such difference is called a common difference of the sequence.

That is, an arithmetic progression is a sequence of form x_i = x_1 + (i - 1) d, where d is a common difference of the sequence.

There is a secret list of n integers a_1, a_2, …, a_n.

It is guaranteed that all elements a_1, a_2, …, a_n are between 0 and 10^9, inclusive.

This list is special: if sorted in increasing order, it will form an arithmetic progression with positive common difference (d > 0). For example, the list [14, 24, 9, 19] satisfies this requirement, after sorting it makes a list [9, 14, 19, 24], which can be produced as x_n = 9 + 5 ⋅ (n - 1).

Also you are also given a device, which has a quite discharged battery, thus you can only use it to perform at most 60 queries of following two types:

  * Given a value i (1 ≤ i ≤ n), the device will show the value of the a_i.
  * Given a value x (0 ≤ x ≤ 10^9), the device will return 1 if an element with a value strictly greater than x exists, and it will return 0 otherwise.



Your can use this special device for at most 60 queries. Could you please find out the smallest element and the common difference of the sequence? That is, values x_1 and d in the definition of the arithmetic progression. Note that the array a is not sorted.

Interaction

The interaction starts with a single integer n (2 ≤ n ≤ 10^6), the size of the list of integers.

Then you can make queries of two types:

  * "? i" (1 ≤ i ≤ n) — to get the value of a_i.
  * "> x" (0 ≤ x ≤ 10^9) — to check whether there exists an element greater than x



After the query read its result r as an integer.

  * For the first query type, the r satisfies 0 ≤ r ≤ 10^9. 
  * For the second query type, the r is either 0 or 1.
  * In case you make more than 60 queries or violated the number range in the queries, you will get a r = -1.
  * If you terminate after receiving the -1, you will get the "Wrong answer" verdict. Otherwise you can get an arbitrary verdict because your solution will continue to read from a closed stream. 



When you find out what the smallest element x_1 and common difference d, print

  * "! x_1 d" 



And quit after that. This query is not counted towards the 60 queries limit.

After printing any query 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.



Hacks

For hack, use the following format:

The first line should contain an integer n (2 ≤ n ≤ 10^6) — the list's size.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — the elements of the list.

Also, after the sorting the list must form an arithmetic progression with positive common difference.

Example

Input


4

0

1

14

24

9

19



Output



&gt; 25

&gt; 15

? 1

? 2

? 3

? 4

! 9 5

Note

Note that the example interaction contains extra empty lines so that it's easier to read. The real interaction doesn't contain any empty lines and you shouldn't print any extra empty lines as well.

The list in the example test is [14, 24, 9, 19].

Solution (in JAVA):

Solution:
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.Closeable;
import java.io.Flushable;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.Iterator;
import java.util.Random;
import java.util.StringTokenizer;

/**
 * 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;
			Input in = new Input(inputStream);
			Output out = new Output(outputStream);
			EArithmeticProgression solver = new EArithmeticProgression();
			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 EArithmeticProgression {
		private final int iinf = 1_000_000_000;

		public EArithmeticProgression() {
		}

		public void solve(int kase, InputReader in, Output pw) {
			int n = in.nextInt(), mval;
			{
				int l = 0, r = iinf;
				while(l<r) {
					int mid = (l+r)/2;
					pw.println("> "+mid);
					pw.flush();
					if(in.nextInt()==1) {
						l = mid+1;
					}else {
						r = mid;
					}
				}
				mval = l;
			}
			HashSet<Integer> valid = new HashSet<>(n);
			for(int i = 1; i<=n; i++) {
				pw.println("? "+i);
				pw.flush();
				int x = in.nextInt();
				if(x!=mval) {
					x = mval-x;
					for(int j = 1; j<n; j++) {
						if(x%j==0) {
							valid.add(x/j);
						}
					}
					break;
				}
			}
			ArrayList<Integer> left = new ArrayList<>();
			for(int i = 1; i<=n; i++) {
				left.add(i);
			}
			Random rand = new Random();
			while(valid.size()>1) {
				int q = rand.nextInt(left.size());
				pw.println("? "+left.get(q));
				pw.flush();
				left.remove(q);
				HashSet<Integer> next = new HashSet<>(n);
				int x = in.nextInt();
				if(x!=mval) {
					x = mval-x;
					for(int j = 1; j<n; j++) {
						if(x%j==0&&valid.contains(x/j)) {
							next.add(x/j);
						}
					}
					valid = next;
				}
			}
			int d = valid.iterator().next();
			pw.println("!", mval-(n-1)*d, d);
			pw.flush();
		}

	}

	static interface InputReader {
		int nextInt();

	}

	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(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(String s) {
			sb.append(s);
			println();
		}

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

		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 Input implements InputReader {
		BufferedReader br;
		StringTokenizer st;

		public Input(InputStream is) {
			this(is, 1<<20);
		}

		public Input(InputStream is, int bs) {
			br = new BufferedReader(new InputStreamReader(is), bs);
			st = new StringTokenizer("");
		}

		public boolean hasNext() {
			try {
				while(st==null||!st.hasMoreTokens()) {
					String s = br.readLine();
					if(s==null) {
						return false;
					}
					st = new StringTokenizer(s);
				}
				return true;
			}catch(Exception e) {
				return false;
			}
		}

		public String next() {
			if(!hasNext()) {
				throw new InputMismatchException();
			}
			return st.nextToken();
		}

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

	}
}