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.lang.*;
import java.io.*;
import java.math.BigInteger;

public class C
{
	static Scanner in;
	public static void main(String[] args) {
		in = new Scanner(System.in);
		int n = in.nextInt();
		int l=1, r= n;
		int s = query(l, r);
		if(query(1, s) == s) {
			l = 1; r = s-1;
			while(l<r) {
				int m = (l+r+1)/2;
				if(query(m, s) == s) l = m;
				else r = m-1;
			}
			solve(l);
		} else {
			l = s+1; r = n;
			while(l<r) {
				int m = (l+r)/2;
				if(query(s, m) == s) r = m;
				else l = m+1;
			}
			solve(l);
		}
		
	}
	
	static int query(int l, int r) {
		if(l>=r) return -1;
		System.out.println("? " + l + " " + r);
		System.out.flush();
		return in.nextInt();
	}
	
	static void solve(int s) {
		System.out.println("! " + s);
		System.out.flush();
	}
}