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.*; 
import java.math.*; 

public class Main {
	static int right(int max,int start,int end,int n) {
		FastScanner f = new FastScanner(); 
		if(end<start) {
			return n-1;
		}
		else {
			int mid=(start+end)/2;
			System.out.println("? "+(max+1)+" "+(mid+1));
			int place=f.nextInt()-1;
			if(place==max) {
				return Math.min(mid, right(max,start,mid-1,n));
			}
			else {
				return right(max,mid+1,end,n);
			}
		}
		
	}
	static int left(int max,int start,int end) {
//		System.out.println(start+" "+end);
		FastScanner f = new FastScanner(); 
		if(end<start) {
			return 0;
		}
		else {
			int mid=(start+end)/2;
			System.out.println("? "+(mid+1)+" "+(max+1));
			int place=f.nextInt()-1;
			if(place==max) {
				return Math.max(mid, left(max,mid+1,end));
			}
			else {
				return left(max,start,mid-1);
			}
		}
		
	}
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=1;
//		t=f.nextInt();
		PrintWriter out=new PrintWriter(System.out);
		int mod=1000000007;
		while(t>0) {
			t--;
			int n=f.nextInt();
			System.out.println("? "+1+" "+(n));
			int ind=f.nextInt()-1;
			if(ind==(n-1)) {
				System.out.println("! "+(left(ind,0,ind-1)+1));	
			}
			else if(ind==0) {
				System.out.println("! "+(right(ind,ind+1,n-1,n)+1));
			}
			else {
				System.out.println("? "+(ind+1)+" "+(n));
				int place=f.nextInt()-1;
				if(place==ind) {
					System.out.println("! "+(right(ind,ind+1,n-1,n)+1));
				}
				else {
					System.out.println("! "+(left(ind,0,ind-1)+1));	
				}
			}
			
		}
		out.close();
	} 
	static void sort(int [] a) {
        ArrayList<Integer> q = new ArrayList<>();
        for (int i: a) q.add(i);
        Collections.sort(q);
        for (int i = 0; i < a.length; i++) a[i] = q.get(i);
    }
    
	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());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
}