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<iostream>
#include<set>
#include<map>
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
using namespace std;
#include<vector>
#include<queue>
#include<algorithm>
#define ll long long 
#include<climits>
#define read(i,l,r) for(i=l;i<r;i++) 
#include<math.h>
map<pair<int,int>,int> mp;
int get(int l,int r){
	if(mp.count({l,r})){
		return mp[{l,r}];
	}
	cout<<"? "<<l<<" "<<r<<endl;
	int pos;
	cin>>pos;
	mp[{l,r}]=pos;
	return pos;
}
void put(int n){
	cout<<"! "<<n<<endl;
}


void solve(int testcase){
	int n,check;
	cin>>n;
	
	int pos=get(1,n);
	if(pos==1)
		check=-1;
	
	else 
		check=get(1,pos);

	
	if(check==pos){
		int l=1,r=pos-1;
		while(l<=r){
			int mid=l+(r-l)/2;
			if(get(mid,pos)==pos){
				l=mid+1;
			}
			else{
				r=mid-1;
			}
		}
		put(l-1);
	}
	else{
		int l=pos+1,r=n;
		while(l<=r){
			int mid=l+(r-l)/2;
			if(get(pos,mid)==pos){
				r=mid-1;
				
			}
			else l=mid+1;
		}
		put(r+1);
	}
		
}


int main(){
	fast;
	int t=1;
//	cin>>t;
	for(int i=1;i<=t;i++){
		solve(i);
	}
}