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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Khater
 */
public class HH {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        C1GuessingTheGreatestEasyVersion solver = new C1GuessingTheGreatestEasyVersion();
        solver.solve(1, in, out);
        out.close();
    }

    static class C1GuessingTheGreatestEasyVersion {
        int idx;
        public void solve(int testNumber, Scanner sc, PrintWriter pw) {
            int t = 1;
            //   t = sc.nextInt();
            while (t-- > 0) {
                int n = sc.nextInt();
                 idx = ask(1, n, sc, pw);
                 int b = ask(1,idx,sc,pw);
                int ans =1;
                if(b==idx) {
                	 ans = bbs(1,idx,sc,pw);
                }else {
                	 ans = abs(idx,n,sc,pw);
                }
                pw.println("! "+ans);
                
            }

        }

        int abs(int l, int r, Scanner sc, PrintWriter pw) {
            if (l >= r-1) {
               if(ask(idx, l, sc, pw)!=idx)
                return r;
               else return l;
            }
            
            int mid = (l+r)/2;
            int b = ask(idx, mid, sc, pw);
             if(b==idx)return abs(l,mid,sc,pw);
             else return abs(mid+1,r,sc,pw);
        }
        int bbs(int l, int r, Scanner sc, PrintWriter pw) {
        	
            if (l >= r-1) {
               if(ask(r, idx, sc, pw)==idx)
                return r;
               else return l;
            }
            int mid = (l+r)/2;
            int a = ask(mid, idx, sc, pw);
             if(a==idx)return bbs(mid,r,sc,pw);
             else return bbs(l,mid-1,sc,pw);
        }
        
        int ask(int l, int r, Scanner sc, PrintWriter pw) {
        	if(l>=r) return -1;
            pw.println("? " + l + " " + r);
            pw.flush();
            return sc.nextInt();

        }

    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

    }
}