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 CPP):

Solution:
#include <cstdio>

int sMax;

int getMid(int p1, int p2)
{
	return p1 + ((p2 - p1) / 2);
}

int getPos(int p1, int p2)
{
	int p;
	
	printf("? %d %d\n", p1, p2);
	fflush(stdout);
	
	scanf("%d", &p);
	
	return p;
}

int binarySearch1(int p1, int p2)
{
	if(p1 == p2)
		return p1;
	
	int mid = getMid(p1, p2);
	int pos = getPos(mid + 1, sMax);
	
	if(pos == sMax)
		return binarySearch1(mid + 1, p2);
	
	else
		return binarySearch1(p1, mid);
}

int binarySearch2(int p1, int p2)
{
	if(p1 == p2)
		return p1;
	
	int mid = getMid(p1, p2);
	int pos = getPos(sMax, mid);
	
	if(pos == sMax)
		return binarySearch2(p1, mid);
	
	else
		return binarySearch2(mid + 1, p2);
}

int getSolution(int p1, int p2)
{
	int pos1;
	int pos2;
	
	pos1 = getPos(p1, p2);
	sMax = pos1;
	
	if(pos1 == p1)
		return binarySearch2(pos1 + 1, p2);
	
	if(pos2 == p2)
		return binarySearch1(p1, pos1 - 1);
	
	pos2 = getPos(p1, pos1);
	
	if(pos1 == pos2)
		return binarySearch1(p1, pos1 - 1);
	
	else
		return binarySearch2(pos1 + 1, p2);
}

int main()
{
	int n;
	
	scanf("%d", &n);
	
	printf("! %d\n", getSolution(1, n));
	
	fflush(stdout);
}