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 static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
 
public class Template {

	public static void main(String[] args) throws Exception{
		BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(infile.readLine());
        int N = Integer.parseInt(st.nextToken());
        System.out.println("? 1 " + N);
        System.out.flush();
        int secondLarge = Integer.parseInt(infile.readLine());
        int ans = 0;
        if(secondLarge > 1) {
        	// check left then
        	System.out.println("? 1 " + secondLarge);
            System.out.flush();
            int index = Integer.parseInt(infile.readLine());
            if(index == secondLarge) {
            	// definitely we check left one
            	int start = 1;
            	int end = secondLarge- 1;
            	while(start < end) {
            		int mid = (start+end+1)/2;
            		System.out.println("? " + mid + " " + secondLarge);
                    System.out.flush();
                    index = Integer.parseInt(infile.readLine());
                    if(index == secondLarge) {
                    	start = mid;
                    }
                    else {
                    	end = mid-1;
                    }
            	}
            	ans = start;
            }
        }
        
        if(ans == 0 && secondLarge < N) {
        	// then go for right
        	int start = secondLarge+1;
        	int end = N;
        	while(start < end) {
        		int mid = (start+end)/2;
        		System.out.println("? " + secondLarge + " " + mid);
                System.out.flush();
                int index = Integer.parseInt(infile.readLine());
                if(index == secondLarge) {
                	end = mid;
                }
                else {
                	start = mid+1;
                }
        	}
        	ans = start;
        }
        
        System.out.println("! " + ans);
        System.out.flush();
        
        
	}
	
	 public static int[] readArr(int N, BufferedReader infile, StringTokenizer st) throws Exception
	    {
	        int[] arr = new int[N];
	        st = new StringTokenizer(infile.readLine());
	        for(int i=0; i < N; i++)
	            arr[i] = Integer.parseInt(st.nextToken());
	        return arr;
	    }
	    
	    
}