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.*;
import java.util.*;
import java.util.stream.*;

public class E {
	public static boolean DEBUG = false;

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		StringTokenizer st;

		int n = nextInt(br);
		int queries = 60;

		long lo = 0;
		long hi = 1000000000;

		while (lo < hi) {
			long mid = (lo + hi + 1) / 2;

			queries--;
			pw.printf("> %d\n", mid);
			pw.flush();

			if (nextInt(br) == 1) {
				lo = mid;
			} else {
				hi = mid - 1;
			}
		}

		// debug(lo+1);

		long highest = lo+1;

		int[] picks = new int[n];

		for (int i = 0; i < n; i++) {
			picks[i] = i+1;
		}

		for (int i = n-1; i >= 0; i--) {
			int s = (int)Math.floor(Math.random() * i);

			int t = picks[s];
			picks[s] = picks[i];
			picks[i] = t;
		}

		ArrayList<Integer> v = new ArrayList<Integer>();

		for (int i = 0; i < n && i < queries; i++) {
			pw.printf("? %d\n", picks[i]);
			pw.flush();

			v.add(nextInt(br));
		}

		long d = 0;

		for (int i = 0; i < v.size(); i++) {
			for (int j = i+1; j < v.size(); j++) {
				// debug(v.get(i), v.get(j), Math.abs(v.get(j) - v.get(i)));
				if (d == 0) {
					d = Math.abs(v.get(j) - v.get(i));
				} else {
					d = gcd(d, Math.abs(v.get(j) - v.get(i)));
				}
			}

			if (picks[i] != highest) {
				// debug(highest, v.get(i), Math.abs(highest - v.get(i)));
				if (d == 0) {
					d = Math.abs(highest - v.get(i));
				} else {
					d = gcd(d, Math.abs(highest - v.get(i)));
				}
			}
		}

		long a = highest - (n-1) * d;

		pw.printf("! %d %d\n", a, d);

		br.close();
		pw.close();
	}

	public static long gcd(long a, long b) {
		if (a > b) {
			return gcd(b, a);
		}

		if (a == 0) {
			return b;
		}

		return gcd(b % a, a);
	}

	// -----------------------------------------------------------------------------------------------------------------
	// -----------------------------------------------------------------------------------------------------------------
	// -----------------------------------------------------------------------------------------------------------------

	public static void debug(Object obj, String end) {
		if (DEBUG) {
			if (obj instanceof boolean[]) {
				System.out.print(Arrays.toString((boolean[])obj));
			} else if (obj instanceof byte[]) {
				System.out.print(Arrays.toString((byte[])obj));
			} else if (obj instanceof short[]) {
				System.out.print(Arrays.toString((short[])obj));
			} else if (obj instanceof char[]) {
				System.out.print(Arrays.toString((char[])obj));
			} else if (obj instanceof int[]) {
				System.out.print(Arrays.toString((int[])obj));
			} else if (obj instanceof long[]) {
				System.out.print(Arrays.toString((long[])obj));
			} else if (obj instanceof float[]) {
				System.out.print(Arrays.toString((float[])obj));
			} else if (obj instanceof double[]) {
				System.out.print(Arrays.toString((double[])obj));
			} else if(obj instanceof Object[]) {
				debug((Object[])obj);
			} else {
				System.out.print(obj);
			}
			System.out.print(end);
		}
	}

	public static void debug(Object... args) {
		if (DEBUG) {
			System.out.print("#");
			for (int i = 0; i < args.length; i++) {
				debug(args[i], " ");
			}
			System.out.println();
		}
	}

	public static void debug(Suspended sus) {
		if (DEBUG) {
			debug(sus.eval());
		}
	}

	public static void debugGrid(int[][] grid) {
		if (DEBUG) {
			for (int i = 0; i < grid.length; i++) {
				System.out.print("#");
				for (int j = 0; j < grid[0].length; j++) {
					System.out.print(String.format("%3d ", grid[i][j]));
				}
				System.out.println();
			}
		}
	}

	public static StringTokenizer getst(BufferedReader br) throws Exception {
		return new StringTokenizer(br.readLine(), " ");
	}

	public static int nextInt(BufferedReader br) throws Exception {
		return Integer.parseInt(br.readLine());
	}

	public static int nextInt(StringTokenizer st) throws Exception {
		return Integer.parseInt(st.nextToken());
	}

	public static long nextLong(BufferedReader br) throws Exception {
		return Long.parseLong(br.readLine());
	}

	public static long nextLong(StringTokenizer st) throws Exception {
		return Long.parseLong(st.nextToken());
	}

	public static double nextDouble(BufferedReader br) throws Exception {
		return Double.parseDouble(br.readLine());
	}

	public static double nextDouble(StringTokenizer st) throws Exception {
		return Double.parseDouble(st.nextToken());
	}
}

interface Suspended {
	public Object eval();
}