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:
/// #459
#include <bits/stdc++.h>
using namespace std;
typedef pair<int,int > pii;
typedef long long int lld;
#define pi acos(-1)
#define fr(i,m,n) for(i=m;i<n;i++)
#define fu(i,m,n) for(i=m;i>=n;i--)
#define vec vector<lld>
#define pb push_back
#define pp pop_back()
#define ft first
#define sd second
#define all(v) v.begin(),v.end()
#define mom(ara) memset(ara,0,sizeof(ara));
#define m1m(ara) memset(ara,-1,sizeof(ara));
#define endl "\n"
#define eps 1.19209e-07
lld query(lld l,lld r)
{
	lld x;
	cout<<"? "<<l<<" "<<r<<endl;
	cout.flush();
	cin>>x;
	return x;
}
int main()
{ios_base::sync_with_stdio(0);cout.tie(0);cin.tie(0);
    lld i,j,k,a,b,c,x,y,z,n,m,ans,t;
    cin>>n;
    z=query(1,n);
    if(z!=1) x=query(1,z);
    else x=0;
    if(x==z)
    {
    	lld l=1,r=z-1;
    	while(l<r)
    	{
    		lld mid=(l+r)/2;
    		if(query(mid+1,z)==z)
    		{
    			l=mid+1;
    		}
    		else r=mid;
    	}
    	cout<<"! "<<l<<endl;
    	cout.flush();
    }
    else
    {
    	lld l=z+1,r=n;
    	while(l<r)
    	{
    		lld mid=(l+r)/2;
    		if(query(z,mid)==z)
    		{
    			r=mid;
    		}
    		else l=mid+1;
    	}
    	cout<<"! "<<l<<endl;
    	cout.flush();
    }
}