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.Scanner;


public class Main {
    public static Scanner scammer = new Scanner(System.in);
    public static int querry(int left, int right, boolean rite){
        if(left == right)
            return rite ? right + 1 : left - 1;

        System.out.print("? ");
        System.out.print(left);
        System.out.print(" ");
        System.out.println(right);
        System.out.flush();
        return scammer.nextInt();
    }
    public static void solve(boolean right, int smax, int n){
        int p = right ? smax : 1, k = right ? n : smax, mid = (p + k) >> 1;
        while(p <= k){
            mid = (p + k) >> 1;
            int outcome = right ? querry(smax, mid, true) : querry(mid, smax, false);
            if(outcome == smax){
                if(right)
                    k = mid - 1;
                else
                    p = mid + 1;
            }
            else{
                if(right)
                    p = ++mid;
                else
                    k = --mid;
            }
        }
        System.out.print("! ");
        System.out.println(mid);
        System.out.flush();
    }

    public static void solve_small(int n){
        int x = querry(1, n, false);
        System.out.printf("! %d\n", x == 1 ? 2 : 1);
        System.out.flush();
    }

    public static void main(String[] args) {
        int n = scammer.nextInt();
        int smax = querry(1, n, false);
        if(n < 3) {
            solve_small(n);
            return;
        }
        boolean right = false;
        int drugi_max = querry(1, smax, false);
        if(drugi_max != smax)
            right = true;
        solve(right, smax, n);
    }
}