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.util.StringTokenizer;

import static java.lang.System.in;
import static java.lang.System.out;

public class Main {
    public static void main(String[] args){
        FastReader fr=new FastReader();
        int tc=1;
        while(tc-->0){
            int n=fr.nextInt();

            int sm=query(1,n,fr);
            if(sm!=1 && query(1,sm,fr)==sm){
                int l=1,r=sm-1;
                while(l<=r){
                    int mid=(l+r)/2;
                    if(query(mid,sm,fr)==sm){
                        l=mid+1;
                    }else{
                        r=mid-1;
                    }
                }
                out.println("! "+r);
            }else{
                int l=sm+1,r=n;
                while(l<=r){
                    int mid=(l+r)/2;
                    if(query(sm,mid,fr)==sm){
                        r=mid-1;
                    }else{
                        l=mid+1;
                    }
                }
                out.println("! "+l);
            }
            out.flush();
        }
    }

    static int query(int l,int r,FastReader fr){
        out.println("? "+l+" "+r);
        out.flush();
        return fr.nextInt();
    }

    static class FastReader{
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(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());
        }

        byte nextByte(){
            return Byte.parseByte(next());
        }

        Float nextFloat(){
            return Float.parseFloat(next());
        }

        long nextLong(){
            return Long.parseLong(next());
        }

        double nextDouble(){
            return Double.parseDouble(next());
        }

        short nextShort(){
            return Short.parseShort(next());
        }

        Boolean nextbol(){
            return Boolean.parseBoolean(next());
        }

        String nextLine(){
            String str = "";
            try{
                str = br.readLine();
            }
            catch (IOException e){
                e.printStackTrace();
            }
            return str;
        }
    }
}