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.awt.*;
public class C
{
    BufferedReader in;
    PrintWriter ob;
    StringTokenizer st;
  
    public static void main(String[] args) throws IOException {
        new C().run();
    }
    void run() throws IOException {
        in=new BufferedReader(new InputStreamReader(System.in));
        ob=new PrintWriter(System.out);
        solve();
        System.out.flush();
    }

    void solve() throws IOException {
        int n = ni();

        System.out.println("? 1 "+n);
        int s_max_pos = ni();

        int left_sec_max = 0;

        if( s_max_pos != 1 ) {
          System.out.println("? 1"+" "+s_max_pos);
          left_sec_max = ni();
        }

        if( s_max_pos != 1 && left_sec_max == s_max_pos ) {

            int left = 1, right = s_max_pos;
            while( right - left > 1 ) {
                int mid = left + (right - left)/2;

                System.out.println("? "+mid+" "+s_max_pos);
                int res = ni();
                if ( res == s_max_pos )
                  left = mid;
                else
                  right = mid - 1;
            }

            if( right == s_max_pos ) {
                System.out.println("! "+left);
                return;
            }

            System.out.println("? "+right+" "+s_max_pos);
            int f = ni();

            if( f == s_max_pos )
              System.out.println("! "+right);
            else
              System.out.println("! "+left);
        } else {

          int left = s_max_pos, right = n;
            while( right - left > 1 ) {
                int mid = left + (right - left)/2;

                System.out.println("? "+s_max_pos+" "+mid);
                int res = ni();
                if ( res == s_max_pos )
                  right = mid;
                else
                  left = mid + 1;
            }

            if( left == s_max_pos ) {
                System.out.println("! "+right);
                return;
            }

            System.out.println("? "+s_max_pos+" "+left);
            int f = ni();

            if( f == s_max_pos )
              System.out.println("! "+left);
            else
              System.out.println("! "+right);

        }
    }
 
 
    String ns() throws IOException {
        return nextToken();
    }
    long nl() throws IOException {
        return Long.parseLong(nextToken());
    }
    int ni() throws IOException {
        return Integer.parseInt(nextToken());
    }
    double nd() throws IOException {
        return Double.parseDouble(nextToken());
    } 
    String nextToken() throws IOException {
        if(st==null || !st.hasMoreTokens())
            st=new StringTokenizer(in.readLine());
        return st.nextToken();
    }
    int[] nia(int start,int b) throws IOException {
        int a[]=new int[b];
        for(int i=start;i<b;i++)
            a[i]=ni();
        return a;
    }
    long[] nla(int start,int n) throws IOException {
        long a[]=new long[n];
        for (int i=start; i<n ;i++ ) {
            a[i]=nl();
        }
        return a;
    }
    
    public void tr(Object... o) {
        ob.println(Arrays.deepToString(o));
    }
}