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:
//created by Whiplash99
import java.io.*;
import java.util.*;
public class C
{
    private static int count;
    private static void flush(){System.out.flush();}
    private static void ask(int l, int r) throws Exception
    {
        count++; if(count>20) throw new Exception();
        System.out.println("? "+l+" "+r);
        flush();
    }
    private static void answer(int x)
    {
        System.out.println("! "+x);
        flush();
    }
    public static void main(String[] args) throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

        int i,N; count=0;

        N=Integer.parseInt(br.readLine().trim());

        ask(1,N);
        int p=Integer.parseInt(br.readLine().trim());

        boolean flag=true; int ans=0;
        if(p>1)
        {
            ask(1,p); ans=p-1;
            int tmp=Integer.parseInt(br.readLine().trim());
            if(tmp==p)
            {
                flag=false;

                int l=1,r=p-1,mid;
                while (l<=r)
                {
                    mid=(l+r)/2;

                    ask(mid,p);
                    tmp=Integer.parseInt(br.readLine().trim());

                    if(tmp==p)
                    {
                        ans=mid;
                        l=mid+1;
                    }
                    else r=mid-1;
                }
            }
        }
        if(flag)
        {
            int l=p+1,r=N,mid; ans=p+1;
            while (l<=r)
            {
                mid=(l+r)/2;

                ask(p,mid);
                int tmp=Integer.parseInt(br.readLine().trim());

                if(tmp==p)
                {
                    ans=mid;
                    r=mid-1;
                }
                else l=mid+1;
            }
        }

        answer(ans);
    }
}