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.util.*;
import java.io.*;
 
public class Main2 {
	static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
 
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
 
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
 
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
 
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
 
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
 
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
	static FastReader sc = new FastReader();
	static int n;
	public static void main (String[] args) {
		PrintWriter out = new PrintWriter(System.out);
		int t = 1;
//    	t = sc.nextInt();
    	 while(t-->0) {
       		n = sc.nextInt();
       		int sm = ask(1,n);
       		int left = ask(1,sm);
       		int ans = -1;
       		if(left==sm) {
       			int l=1, h=sm-1;
       			while(l<=h) {
       				int mid = (l+h)/2;
       				left = ask(mid,sm);
       				if(left == sm) {
       					ans = mid;
       					l = mid+1;
       				}
       				else h = mid-1;
       			}
       		}
       		else {
       			int l=sm+1, h=n;
       			while(l<=h) {
       				int mid = (l+h)/2;
       				left = ask(sm,mid);
       				if(left == sm) {
       					ans = mid;
       					h = mid-1;
       				}
       				else l = mid+1;
       			}
       		}
       		System.out.println("! "+ans);
       	}
		out.close();
	}
	
	private static int ask(int idx, int h) {
		if(idx>n || h>n || idx == 0 || h == 0) return -1;
		if(idx == h) return -1;
		System.out.println("? "+idx+" "+h);
		System.out.flush();
		int ret = sc.nextInt();
		return ret;
	}
    private static long pow(int i, int x) {
		long ans = 1;
		while(x>0) {
			if(x%2==0) {
				i*=i;
				x/=2;
			}
			else {
				ans*=i;
				x--;
			}
		}
//		System.out.println(ans+" ans");
		return ans;
	}
}