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.*;
import java.util.*;

public class Solution
{
  public static void main(String args[]) throws Exception
  {
    BufferedReader Rb = new BufferedReader(new InputStreamReader(System.in));
    int n = Integer.valueOf(Rb.readLine());
    int l = 1, r = n;
    System.out.println("? " + l + " " + r);
    System.out.flush();
    int pos = Integer.valueOf(Rb.readLine());
    System.out.println("? " + (pos!=l?l:pos) + " " + (pos!=l?pos:r));
    System.out.flush();
    int n_pos = Integer.valueOf(Rb.readLine());
    boolean flag = true;
    if(n_pos == pos)
    {
      if(pos == l)
      {l = pos + 1;}
      else
      {r = pos - 1; flag = false;}
    }
    else
    {
      if(pos == l)
      {r = pos - 1; flag = false;}
      else
      {l = pos + 1;}
    }
    int last = -1;
    while(l <= r)
    {
      int m = (l + r) / 2;
      int s = (m < pos)?m:pos;
      int e = (m < pos)?pos:m;
      System.out.println("? " + s + " " + e);
      System.out.flush();
      n_pos = Integer.valueOf(Rb.readLine());
      if(n_pos == pos)
      {
        last = m;
        if(!flag)
        {l = (l + r)/2 + 1;}
        else
        {r = (l + r)/2 - 1;}
      }
      else
      {
        if(flag)
        {l = (l + r)/2 + 1;}
        else
        {r = (l + r)/2 - 1;}
      }
    }
    System.out.println("! " + last);
    System.out.flush();
  }
}