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.*;
public class C {

	static Scanner sc;
	public static void main(String[] args) {
		sc = new Scanner(System.in);
		int n = sc.nextInt();
		System.out.printf("? %d %d\n", 1, n);
		int m = sc.nextInt();
		if(m > 1) {
			System.out.printf("? %d %d\n", 1, m);
			int r = sc.nextInt();
			if(r == m) {
				findmaxleft(m, n);
			}
			else {
				findmaxright(m, n);
			}
		}
		else {
			findmaxright(1, n);
		}

	}
	static void findmaxleft(int m, int n) {
		int a = 1, b = m;
		while(b - a > 1) {
			int c = (a + b)/2;
			System.out.printf("? %d %d\n", c, m);
			int r = sc.nextInt();
			if(r == m) a = c;
			else b = c;
		}
		System.out.printf("! %d\n", a);
	}
	static void findmaxright(int m, int n) {
		int a = m, b = n;
		while(b - a > 1) {
			int c = (a + b)/2;
			System.out.printf("? %d %d\n", m, c);
			int r = sc.nextInt();
			if(r == m) b = c;
			else a = c;
		}
		System.out.printf("! %d\n", b);
	}

}