Problem: The only difference between the easy and the hard version is the limit to the number of queries.

This is an interactive problem.

There is an array a of n different numbers. In one query you can ask the position of the second maximum element in a subsegment a[l..r]. Find the position of the maximum element in the array in no more than 20 queries.

A subsegment a[l..r] is all the elements a_l, a_{l + 1}, ..., a_r. After asking this subsegment you will be given the position of the second maximum from this subsegment in the whole array.

Input

The first line contains a single integer n (2 ≤ n ≤ 10^5) — the number of elements in the array.

Interaction

You can ask queries by printing "? l r" (1 ≤ l < r ≤ n). The answer is the index of the second maximum of all elements a_l, a_{l + 1}, ..., a_r. Array a is fixed beforehand and can't be changed in time of interaction.

You can output the answer by printing "! p", where p is the index of the maximum element in the array.

You can ask no more than 20 queries. Printing the answer doesn't count as a query.

After printing a 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

To make a hack, use the following test format.

In the first line output a single integer n (2 ≤ n ≤ 10^5). In the second line output a permutation of n integers 1 to n. The position of n in the permutation is the position of the maximum

Example

Input


5

3

4



Output


? 1 5

? 4 5

! 1

Note

In the sample suppose a is [5, 1, 4, 2, 3]. So after asking the [1..5] subsegment 4 is second to max value, and it's position is 3. After asking the [4..5] subsegment 2 is second to max value and it's position in the whole array is 4.

Note that there are other arrays a that would produce the same interaction, and the answer for them might be different. Example output is given in purpose of understanding the interaction.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class cf {
	static Reader sc = new Reader();
	// static Scanner sc = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		int n = sc.nextInt();
		int ans = -1;
		int index = query(1, n);
		if (index == 1 || query(1, index) != index) {
			ans = n;
			// binary search
			int l = index, r = n;
			while (l < r) {
				int mid = l + (r - l) / 2;
				if (l + 1 == r) {
					if (l > index && query(index, l) == index) {
						ans = l;
					} else {
						ans = r;
					}
					break;
				}
				if (index == mid)
					break;
				if (query(index, mid) == index) {
					ans = mid;
					r = mid;
				} else {
					l = mid;
				}
			}
		} else {
			ans = 1;
			// binary search
			int l = 1, r = index;
			while (l < r) {
				int mid = l + (r - l) / 2;
				if (l + 1 == r) {
					if (r < index && query(r, index) == index) {
						ans = r;
					} else {
						ans = l;
					}
					break;
				}
				if (index == mid)
					break;
				if (query(mid, index) == index) {
					ans = mid;
					l = mid;
				} else {
					r = mid;
				}
			}
		}
		out.println("! " + ans);
		out.close();
	}

	private static int query(int x, int y) {
		out.println("? " + x + " " + y);
		out.flush();
		return sc.nextInt();
	}

	static class Reader {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
				}
			return st.nextToken();
		}

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

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

}