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 <bits/stdc++.h>
using namespace std;

map<pair<int, int>, int> mp;

int ask(int l, int r)
{
	if (l == r) return -1;
	if (mp.count(make_pair(l, r))) return mp[make_pair(l, r)];
	printf("? %d %d\n", l, r);
	fflush(stdout);
	int x;
	scanf("%d", &x);
	mp[make_pair(l, r)] = x;
	return x;
}

int answer(int x)
{
	printf("! %d\n", x);
	exit(0);
}

void answer(int x, int y)
{
	ask(x, y);
	if (ask(x, y) == x) answer(y);
	else answer(x);
} 

void answer(int x1, int x2, int x3)
{
	int sec = ask(x1, x3);
	if (sec == x1) answer(x2, x3);
	else if (sec == x2)
	{
		if (ask(x1, x2) == x2) answer(x1);
		else answer(x3);
	}
	else answer(x1, x2);
}

void Answer(int l, int r)
{
	if (l == r) answer(l);
	else if (r - l + 1 == 2) answer(l, r);
	else answer(l, l + 1, l + 2);
}

int main()
{
	int n;
	scanf("%d", &n);
	int sec = ask(1, n), x = ask(1, sec);
	int l = sec == x ? 1 : sec, r = sec == x ? sec : n;
	while (r - l + 1 > 2)
	{
		int mid = (l + r) / 2;
		int val = ask(sec == x ? mid : sec, sec == x ? sec : mid);
		if (val == sec)
		{
			if (sec == x) l = mid;
			else r = mid;
		}
		else
		{
			if (sec == x) r = mid;
			else l = mid;
		}
	}
	Answer(l, r);
	return 0;
}