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.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class C{
	
	static FastScanner fs = new FastScanner();
	static PrintWriter out = new PrintWriter(System.out);
		
	

	
    public static void main(String[] args) throws IOException {
    	
		
//    	FastScanner fs = new FastScanner();
//    	PrintWriter out = new PrintWriter(System.out);
  		
    	int tt = 1;
    	while(tt-->0) {
    		
    		int n = fs.nextInt();
    		
    		int ind = query(1, n);
    		int l = -1, r = -1;
    		
    		boolean inleft = false;
    		
    		if(ind==1) {
    			l = 2;
    			r = n;
    			inleft = false;
    		}
    		else if(ind==n) {
    			l = 1;
    			r = n-1;
    			inleft = true;
    		}
    		else {
    			int q = query(1, ind);
    			if(q==ind) {
    				l = 1; r = ind-1;
    				inleft = true;
    			}
    			else {
    				l = ind + 1;
    				r = n;
    				inleft = false;
    			}
    		}
    		
    		
    		
    		if(inleft) {
    			
    			while(l<r) {
    				int mid = (l+r+1)/2;
    				int k = query(mid, ind);
    				if(k==ind) {
    					l = mid;
    				}
    				else {
    					r = mid - 1;
    				}
    			}
    			
    			output(l);
    			
    		}
    		
    		else {
    			
    			while(l<r) {
    				int mid = (l+r)/2;
    				int k = query(ind, mid);
    				if(k==ind) {
    					r = mid;
    				}
    				else {
    					l = mid + 1;
    				}
    			}
    			
    			output(l);
    			
    			
    			
    			
    		}
    		
    		
    		
    		

    	
    		
    		
    		
    		
    		
    		
    		
    		
    	}
    	
    	out.close();
    		
    	
    	
    }
    
    
    static int query(int l, int r) {
    	out.println("? "+l+" "+r);
    	out.flush();
    	int x = fs.nextInt();
    	return x;
    }
    
    static void output(int x) {
    	out.println("! "+x);
    	out.flush();
    	return;
    }
    
   
    
    
    
    
    
    
    
    static final Random random=new Random();
    
    static <T> void shuffle(T[] arr) {
    	int n = arr.length;
    	for(int i=0;i<n;i++	) {
    		int k = random.nextInt(n);
    		T temp = arr[k]; arr[k] = arr[i]; arr[i] = temp;
    	}
    }
    
    	
    static void ruffleSort(int[] a) {
    	int n=a.length;//shuffle, then sort 
    	for (int i=0; i<n; i++) {
    		int oi=random.nextInt(n); int temp=a[oi];
    		a[oi]=a[i]; a[i]=temp;
    	}
    	Arrays.sort(a);
    }
    
    static void ruffleSort(long[] a) {
    	int n=a.length;//shuffle, then sort 
    	for (int i=0; i<n; i++) {
    		int oi=random.nextInt(n); long temp=a[oi];
    		a[oi]=a[i]; a[i]=temp;
    	}
    	Arrays.sort(a);
    }
   
  
    
    static void reverse(int[] arr, int l, int r) {
    	for(int i=l;i<l+(r-l)/2;i++){
    		int temp = arr[i]; arr[i] = arr[r-i+l-1]; arr[r-i+l-1] = temp;
    	}
    }
    
    static void reverse(long[] arr, int l, int r) {
    	for(int i=l;i<l+(r-l)/2;i++){
    		long temp = arr[i]; arr[i] = arr[r-i+l-1]; arr[r-i+l-1] = temp;
    	}
    }
    
    
    static <T> void reverse(T[] arr, int l, int r) {
    	for(int i=l;i<l+(r-l)/2;i++) {
    		T temp = arr[i]; arr[i] = arr[r-i+l-1]; arr[r-i+l-1] = temp;
    	}
    }
    	
    	
    static class FastScanner{
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
     
    	public String next(){
    		while(!st.hasMoreElements()){
    			try{
    				st = new StringTokenizer(br.readLine());
    			} catch(IOException e){
    				e.printStackTrace();
    			}
    		}
    		return st.nextToken();
    	}
    		
    	public String nextLine() throws IOException {
    		return br.readLine();
    	}
    		
    	public int nextInt(){
    		return Integer.parseInt(next());
    	}
     
    	public int[] readArray(int n){
    		int[] a = new int[n];
    		for(int i=0;i<n;i++)
    			a[i] = nextInt();
    		return a;
    	}
    		
    	public long nextLong() {
    		return Long.parseLong(next());
    	}
    		
    	public char nextChar() {
    		return next().toCharArray()[0];
    	}
    }
   	
}