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.*;
import java.text.*;
public class Main  implements Runnable {
    FastScanner sc;
    PrintWriter pw;
    
    final class FastScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner() {
            try {
                br = new BufferedReader(new InputStreamReader(System.in));
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
 
        public long nlo() {
            return Long.parseLong(next());
        }
 
        public String next() {
            if (st.hasMoreTokens()) return st.nextToken();
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                e.printStackTrace();
            }
            return st.nextToken();
        }
 
        public int ni() {
            return Integer.parseInt(next());
        }
 
        public String nli() {
            String line = "";
            if (st.hasMoreTokens()) line = st.nextToken();
            else try {
                return br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            while (st.hasMoreTokens()) line += " " + st.nextToken();
            return line;
        }
 
        public double nd() {
            return Double.parseDouble(next());
        }
    }
    public static void main(String[] args) throws Exception
    {
        new Thread(null,new Main(),"codeforces",1<<28).start();
    }
    public void run()
    {
       // sc=new FastScanner();
        //pw=new PrintWriter(System.out);
        try{
            solve();
        }
        catch(Exception e)
        {
            ;
           // pw.println(e);
        }
       //pw.flush();
        //pw.close();
    }
    public long gcd(long a,long b)
    {
        return b==0L?a:gcd(b,a%b);
    }
    public long ppow(long a,long b,long mod)
    {
        if(b==0L)
        return 1L;
        long tmp=1;
        while(b>1L)
        {
            if((b&1L)==1L)
            tmp*=a;
            a*=a;
            a%=mod;
            tmp%=mod;
            b>>=1;
        }
        return (tmp*a)%mod;
    }
    public long pow(long a,long b)
    {
        if(b==0L)
        return 1L;
        long tmp=1;
        while(b>1L)
        {
            if((b&1L)==1)
            tmp*=a;
            a*=a;
            b>>=1;
        }
        return (tmp*a);
    }
    public  int gcd(int x,int y)
    {
        return y==0?x:gcd(y,x%y);
    }
   
    //////////////////////////////////
    /////////////  LOGIC  ///////////
    ////////////////////////////////
    public void solve() throws Exception{
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        System.out.println("? "+1+" "+n);
        int m=sc.nextInt();
        if(m!=1&&m!=n){
            System.out.println("? 1 "+m);
            int x=sc.nextInt();
            if(x==m)
                System.out.println("! "+f(1,m,sc));
            else
                System.out.println("! "+g(m,n,sc));
        }
        else if(m==1)
            System.out.println("! "+g(1,n,sc));
        else
            System.out.println("! "+f(1,n,sc));
    }
    public int f(int s,int e,Scanner sc){
        int ans=s;
        int val=e;
        e--;
        while(s<=e){
            int m=(s+e)/2;
            System.out.println("? "+m+" "+val);
            int x=sc.nextInt();
            if(x==val){
                ans=m;
                s=m+1;
            }
            else
                e=m-1;
        }
        return ans;
    }
    public int g(int s,int e,Scanner sc){
        int ans=s;
        int val=s;
        s++;
        while(s<=e){
            int m=(s+e)/2;
            System.out.println("? "+val+" "+m);
            int x=sc.nextInt();
            if(x==val){
                ans=m;
                e=m-1;
            }
            else
                s=m+1;
        }
        return ans;
    }
}