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.

Source: codeforces | Difficulty: HARD
Tags: Sorting


Solutions:
from __future__ import print_function
from collections import Counter, OrderedDict
from itertools import permutations as perm
from fractions import Fraction
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
g = lambda : stdin.readline().strip()
gl = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1000000000.0) + 7
inf = float('inf')
(n,) = gil()
dp = {}

def query(l, r):
	if r == l:
		return 0
	if (l, r) not in dp:
		print('?', l, r, flush=True)
		dp[l, r] = gil()[0]
	return dp[l, r]
i = query(1, n)
if query(1, i) == i:
	(l, r) = (1, i)
	ans = i - 1
	while l <= r:
		mid = (l + r) // 2
		if query(mid, i) == i:
			ans = mid
			l = mid + 1
		else:
			r = mid - 1
	print('!', ans, flush=True)
else:
	(l, r) = (i, n)
	ans = i + 1
	while l <= r:
		mid = (l + r) // 2
		if query(i, mid) == i:
			ans = mid
			r = mid - 1
		else:
			l = mid + 1
	print('!', ans, flush=True)


from sys import stdin, stdout

def ask(l, r):
	stdout.write('? ' + str(l) + ' ' + str(r) + '\n')
	stdout.flush()
	nd = int(stdin.readline())
	return nd
try:
	n = int(stdin.readline())
	l = 1
	r = n
	nd = ask(l, r)
	if nd > 1 and ask(1, nd) == nd:
		l = 1
		r = nd
		while l + 1 != r:
			md = (l + r) // 2
			if ask(md, nd) == nd:
				l = md
			else:
				r = md
		stdout.write('! ' + str(l) + '\n')
	else:
		l = nd
		r = n
		while l + 1 != r:
			md = (l + r) // 2
			if ask(nd, md) == nd:
				r = md
			else:
				l = md
		stdout.write('! ' + str(r) + '\n')
except:
	pass