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.*;
import java.util.*;

public class tank {

    //static StringBuilder IO = new StringBuilder();

    //static final Scanner scan = new Scanner(System.in);

    static final FastScanner fs = new FastScanner();

    public static void main(String[] args) {
//        int t = fs.nextInt();
//        while(t-->0){
//            SLV();
//        }

        SLV();
        //System.out.println(IO);
    }

    static void SLV(){
        int n = fs.nextInt();

        int smax = ask(1, n), ans;

        if(smax == 1) {
           ans = fn(2, n, 1);
        }else if(smax == n){
            ans = fn(1, n-1, n);
        }else{
            if(ask(1, smax) == smax){
                ans = fn(1, smax - 1, smax);
            }else ans = fn(smax + 1, n, smax);
        }

        System.out.println("! " + ans);
    }

    static int fn(int l ,int r, int smax){
        int mid = (l + r)/2;

        if(r == l) return l;

        if(smax > r){
            if(l == r-1){
                if(ask(r, smax) == smax) return r;
                else return l;
            }
            if(ask(mid , smax) == smax) return fn(mid, r, smax);
            else return fn(l, mid - 1, smax);
        }else{
            if(l == r - 1){
                if(ask(smax, l) == smax) return l;
                else return r;
            }
            if(smax == ask(smax, mid)) return fn(l, mid, smax);
            else return fn(mid + 1,r, smax);
        }
    }

    static int ask(int l, int r){
        if(l == r) return l;
        System.out.println("? " + l + " " + r);
        int num = fs.nextInt();
        System.out.flush();
        return num;
    }

    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

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

        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }

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