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 C {

	public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int low = 1, high = n;
        out.println("? "+low+" "+high); out.flush();
        int index = in.nextInt();                    
        int left = 0;
        if(index!=1){
        	out.println("? "+low+" "+index); out.flush();
        	left = in.nextInt();
        }        
        if(n==2){
        	if(left==0) out.println("! "+n);
        	else out.println("! "+1);
    	} 
        else if(left==index){
        	low = 1; high = index-1;
        	while(low<=high){
        		int mid = (low+high)/2;
        		out.println("? "+mid+" "+index); out.flush();
        		int x = in.nextInt();
        		if(x==index) low = mid+1;
        		else high = mid-1;	
        	}
        	out.println("! "+high);
        }
        else{
        	low = index+1; high = n;
        	while(low<=high){
        		int mid = (low+high)/2;
        		out.println("? "+index+" "+mid); out.flush();
        		int x = in.nextInt();
        		if(x==index) high = mid-1;
        		else low = mid+1;	
        	}
        	out.println("! "+low);
        }
       	out.flush();
	}
	
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		String next() {
			while(!st.hasMoreTokens())
				try { st = new StringTokenizer(br.readLine()); }
				catch(IOException e) {}
			return st.nextToken();
		}
		
		String nextLine(){
			try{ return br.readLine(); } 
			catch(IOException e) { } return "";
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) {
			int a[] = new int[n];
			for(int i=0;i<n;i++) a[i] = nextInt();
			return a;
		}
	}

	static final Random random = new Random();

	static void ruffleSort(int[] a){
		int n = a.length;
		for(int i=0;i<n;i++){
			int j = random.nextInt(n), temp = a[j];
			a[j] = a[i]; a[i] = temp;
		}
		Arrays.sort(a); 	
	}
}