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:
// Author:   old_school
// Created:  18.02.2021 20:05:30

#include<bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,fma") 
#pragma GCC optimization ("unroll-loops")
using namespace std;
#define lld               long long int
#define ld                long double
#define pb                push_back
#define ppb               pop_back 
#define mp                make_pair
#define F                 first
#define S                 second
#define nl                '\n'
#define all(c)            (c).begin(),(c).end()
#define rall(c)           (c).rbegin(),(c).rend()
#define sz(c)             (c).size()
#define tr(x,a)           for(auto &a : x)
#define psnt(x,a)         (x).find(a)!=(x).end()
#define vpsnt(x,a)        find(all(x),a)!=(x).end()
#define MOD               1000000007
#define tod               1
#define pi                3.1415926536
#define itr(i,a,b)        for(lld i=a;i<=b;i++)
#define itrn(i,a,b)       for(lld i=a;i>=b;i--)
#define iot(n)            for(lld i=0;i<n;i++)
#define pls(n,arr)        lld arr[n]; iot(n)  cin>>arr[i];
#define bye               fflush(stdout)

typedef pair<lld,lld>     pii;
typedef pair<string,lld>  psi;

template <typename T>
bool mycomp(T x,T y){ 
   return (x==y);  //give your condition here
}
bool paircomp(const pair<lld,lld> &x,const pair<lld,lld> &y){ 
return x.second<y.second;
}
void recurse(lld l,lld r,lld sec,lld lim){
	if(l==r) {
		cout<<"! "<<l<<endl;
		return;
	}
	if(r-l==1){
		if(lim==1)
		{
			if(r==sec){
				cout<<"! "<<l<<endl;
				return;
			}
			lld l1;
			cout<<"? "<<r<<" "<<sec<<endl;
			cin>>l1;
			if(l1==sec){
				cout<<"! "<<r<<endl;
			}
			else{
				cout<<"! "<<l<<endl;
			}
		}
		else
		{
			if(l==sec){
				cout<<"! "<<r<<endl;
				return;
			}
		   lld l1;
		   cout<<"? "<<sec<<" "<<l<<endl;
		   cin>>l1;
		   if(l1==sec){
		   	cout<<"! "<<l<<endl;
		   }
		   else{
		       cout<<"! "<<r<<endl;
		   }
		}
		return;
	}
	if(lim==1){lld l1;
		lld md=(l+r)/2;
		cout<<"? "<<md<<" "<<sec<<endl;
		cin>>l1;
		if(l1==sec) recurse(md,r,sec,lim);
		else recurse(l,md-1,sec,lim);
	}
	else{lld l1;
	   lld md=(l+r)/2;
	   cout<<"? "<<sec<<" "<<md<<endl;
		cin>>l1;
		if(l1==sec) recurse(l,md,sec,lim);
		else recurse(md+1,r,sec,lim);
	}
}
void solve(){
   lld n; cin>>n; 
   lld sec;
   cout<<"? "<<1<<" "<<n<<endl;
   cin>>sec;
   if(sec==1) recurse(sec,n,sec,n);
   else if(sec==n) recurse(1,sec,sec,1);
   else{
      lld chk;
      cout<<"? "<<1<<" "<<sec<<endl;
      cin>>chk;
      if(chk==sec) recurse(1,sec,sec,1);
      else recurse(sec,n,sec,n);
  }
} 
int main(){
       ios_base::sync_with_stdio(false);
             cin.tie(0);
            cout.tie(0);
      lld t=1; //cin>>t;  
      while(t--){
            solve();
            cout<<endl;
      }
}