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 middle = Integer.parseInt(infile.readLine());
        int res = 0;
        // check on left
        if(middle > 1) {
        	System.out.println("? 1 " + middle);
        	System.out.flush();
        	int index = Integer.parseInt(infile.readLine());
        	if(index == middle) {
        		int start = 1;
        		int end = middle-1;
        		while(start != end) {
        			int mid = (start+end+1)/2;
        			System.out.println("? " + mid + " " + middle);
        			System.out.flush();
        			index = Integer.parseInt(infile.readLine());
        			if(index == middle) {
        				start = mid;
        			}
        			else {
        				end = mid-1;
        			}
        		}
        		res = start;
        	}
        }
        
//        otherwise check on right
    	if(res == 0 && middle < N) {
    		int start = middle+1;
    		int end = N;
    		while(start != end) {
    			int mid = (start+end)/2;
    			System.out.println("? " + middle + " " + mid);
    			System.out.flush();
    			int index = Integer.parseInt(infile.readLine());
    			if(index == middle) {
    				end = mid;
    			}
    			else {
    				start = mid+1;
    			}
    		}
    		res = start;
    	}
    	System.out.println("! " + res);
    	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;
	    }
}