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.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt();
        System.out.println("? 1 "+n);
        System.out.flush();
        int x=sc.nextInt();
        boolean f=false;
        if (x!=1) {
            f=true;
            System.out.println("? 1 " + x);
        }
        System.out.flush();
        int newX=-1;
        if (f) newX = sc.nextInt();
        if (newX==x){
            int l=1, r=x;
            int ans=-1;
            while (l<=r){

                int m=(l+r)/2;
                if (m==x){
                    System.out.println("! "+(m-1));
                    System.out.flush();
                    return;
                }
                System.out.println("? "+m+" "+x);
                System.out.flush();
                newX=sc.nextInt();
                if (newX==x){
                    ans=m;
                    l=m+1;
                }else {
                    r=m-1;
                }
            }
            System.out.println("! "+(ans));
            System.out.flush();

        }else {
            int r=n,l=x;
            int ans=-1;
            while (l<=r){
                int m=(l+r)/2;
                if (m==x){
                    System.out.println("! "+(m+1));
                    System.out.flush();
                    return;
                }
                System.out.println("? "+x+" "+m);
                System.out.flush();
                newX=sc.nextInt();
                if (newX==x){
                    r=m-1;
                }else {
                    l=m+1;
                }
            }
            System.out.println("! "+(l));
            System.out.flush();

        }
    }
}