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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
 
public class CodeForces {
 

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		FastScanner fs=new FastScanner();
		int n=fs.nextInt();
		int pos=solve(n);
		System.out.println("! "+(pos+1));
//		System.out.println("No of queery = "+count);
 
	}
	private static int solve(int n) {
		int smallx=query(0,n-1);
		if(smallx==0||query(0,smallx)!=smallx) {
			int l=smallx,r=n-1;
			while(r-l>1) {
				int mid=(l+r)/2;
				if(query(smallx,mid)==smallx) {
					r=mid;
				}
				else l=mid;
			}
			return r;
		}
		else {
			int l=0,r=smallx;
			while(r-l>1) {
				int mid=(l+r)/2;
				if(query(mid,smallx)==smallx) {
					l=mid;
				}
				else r=mid;
			}
			return l;
		}
	}
	private static int query(int l,int r) {
		if(l>=r)
			return -1;
		System.out.println("? "+(l+1)+" "+(r+1));
		FastScanner fs=new FastScanner();
		int pos=fs.nextInt();
		return pos-1;
	}
	
	static void mysort(int[] a) {
		//ruffle
		int n=a.length;
		Random r=new Random();
		for (int i=0; i<a.length; i++) {
			int oi=r.nextInt(n), temp=a[i];
			a[i]=a[oi];
			a[oi]=temp;
		}
		
		//then sort
		Arrays.sort(a);
	}
	
	// Use this to input code since it is faster than a Scanner
	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) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
 
}