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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Practice {

	public static void main(String[] args) throws Exception {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		System.out.flush();
		getAns(n, s);

	}

	private static void getAns(int n, Scanner s) {
		// TODO Auto-generated method stub
		
		int sec=call(0,n-1,s);
		if(sec==0||call(0,sec,s)!=sec) {
			int l=sec,r=n-1;
			while(r-l>1) {
				int m=(l+r)/2;
				if(call(sec,m,s)!=sec) {
					l=m;
				}else {
					r=m;
				}
			}System.out.println("! "+(r+1));
		}else {
			int l=0,r=sec;
			while(r-l>1) {
				int m=(l+r)/2;
				if(call(m,sec,s)!=sec) {
					r=m;
				}else {
					l=m;
				}
			}System.out.println("! "+(l+1));
		}
		
	}

	private static int call(int l, int r,Scanner s) {
		// TODO Auto-generated method stub
		System.out.println("? "+(l+1)+" "+(r+1));
		int sec=s.nextInt();
		System.out.flush();
		return sec-1;
	}
}