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:
// c1
import java.util.*;
// import java.lang.*;
import java.io.*;

//           THIS TEMPLATE MADE BY AKSH BANSAL.

public class Solution {
    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;
        }
    }

    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        // ________________________________

        int n= sc.nextInt();
        if(n==1){
            out.println("! 1");
            out.flush();
            return;
        }
        int start = 1, end = n;
        int pos = query(start, end, out, sc);
        if(pos>1 && query(1,pos,out,sc)==pos){
            end = pos-1;
        }else{
            start = pos+1;
        }

        while(start<end){
            int mid = start+(end-start)/2;

            if(start+1==end){
                pos = query(start,end,out,sc);
                if(pos==start){
                    start = end;
                }
                break;
            }
            // checking sub section
            if(pos>=mid){
                    
                int cur = query(mid, pos, out, sc);
                if(cur==pos){
                    start = mid;
                }
                else {
                    end = mid-1;
                }
            }
            else{

                int cur = query(pos, mid, out, sc);
                if(cur==pos){
                    end = mid;
                }
                else {
                    start = mid+1;
                }
            }

        }
        out.println("! "+start);
        out.flush();

    }

    private static int query(int start, int end, PrintWriter out, FastReader sc) {
        out.println("? "+start+" "+end);
        out.flush();
        int res= sc.nextInt();
        return res;
    }
}