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.*;

public class Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    //    long mod = 998244353;
//    long mod = (long)1e9+7;
    long mod = 998244353;
    public static void main(String[] args) throws Exception {
        Main main = new Main();
//        int t = sc.nextInt();
        int t = 1;
        for(int i=1; i<=t; i++){
            main.solve();
        }
        out.flush();
    }

    void solve() throws Exception {
        int n = neInt();
        out.println("? 1 "+n);
        out.flush();
        int secMax = neInt();
        boolean inRight=true;
        if(secMax>1){
            out.println("? 1 "+secMax);
            out.flush();
            int temp = neInt();
            if(temp==secMax){
                inRight = false;
            }
        }
        int co = inRight?1:-1;
        int lo=1,hi;
        if(inRight){
            hi = n-secMax+1;
        } else hi = secMax;
        while(hi>lo+1){
            int mid = (hi+lo)/2;
            int idx = secMax+mid*co-co;
            out.println("? "+Math.min(idx,secMax)+" "+Math.max(idx,secMax));
            out.flush();
            int temp = neInt();
            if(temp==secMax) hi = mid;
            else lo = mid;
        }
        int ans = secMax+hi*co-co;
        out.println("! "+ans);
    }
    String neS(){
        return sc.next();
    }
    int neInt(){
        return Integer.parseInt(sc.next());
    }
    long neLong(){
        return Long.parseLong(sc.next());
    }

    int paIn(String s){return Integer.parseInt(s);}
    public int minimumLengthEncoding(String[] words) {
        int n = words.length;
        Arrays.sort(words, (o1, o2)->(o1.length()-o2.length()));
        int ans = 0;
        for(int i=0; i<n; i++){
            boolean found = false;
            for(int j=i+1; j<n && (!found); j++){
                found |= words[j].endsWith(words[i]);
            }
            if(!found) ans += words[i].length()+1;
        }
        return ans;
    }
}