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.ArrayList;
	import java.util.Arrays;
	import java.util.Scanner;
	import java.io.*; 
	import java.util.*; 

	public class CFEd {
		
		
		public static void main(String[] args) {
			Scanner us = new Scanner(System.in);
			
			int n = us.nextInt();
			
			
			ask(1, n);
			
			int response = us.nextInt();
			if(response == 1) {
				searchRight(response, us, n);
			}
			else if(response == n) {
				searchLeft(response, us);
			}
			else {
				ask(1, response);
				int res2 = us.nextInt();
				
				if(res2 == response) {
					searchLeft(response, us);
				}
				else {
					searchRight(response, us, n);
				}
				
			}
			
			
			
			
			
			
			
			
				
				
		}
		
		
		static void ask(int i, int j) {
			System.out.println("? " + i + " " + j);
			System.out.flush();
		}
		
		
		static void searchLeft(int start, Scanner us) {
			int ubnd = start - 1;
			int lbnd = 1;
			while(ubnd - lbnd > 1) {
				int mid = (ubnd + lbnd)/2;
				ask(mid, start);
				int response = us.nextInt();
				if(response == start) {
					lbnd = mid;
				}
				else {
					ubnd = mid - 1;
				}
			}
			
			if(ubnd == lbnd) {
				System.out.println("! " + ubnd);
			}
			else {
				ask(ubnd, start);
				int response = us.nextInt();
				if(response == start) {
					System.out.println("! " + ubnd);
				}
				else {
					System.out.println("! " + lbnd);
				}
			}
			
			
			
		}
		static void searchRight(int start, Scanner us, int n) {
					int ubnd = n;
					int lbnd = start + 1;
					while(ubnd - lbnd > 1) {
						int mid = (ubnd + lbnd)/2;
						ask(start, mid);
						int response = us.nextInt();
						if(response == start) {
							ubnd = mid;
						}
						else {
							lbnd = mid + 1;
						}
					}
					
					if(ubnd == lbnd) {
						System.out.println("! " + ubnd);
					}
					else {
						
						ask(start, lbnd);
						int response = us.nextInt();
						if(response == start) {
							System.out.println("! " + lbnd);
						}
						else {
							System.out.println("! " + ubnd);
						}
					}
					
					
					
				}
			
			
			
		
		
		
		
	}