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;
#define all(x) x.begin(),x.end()
#define pb push_back
#define ff first
#define ss second
#define fast ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define ar array
#define int long long 
#define F(i,a,b) for(i=a;i<b;++i)
#define RF(i,b,a) for(i=b;i>=a;--i) 
#define NL cout<<"\n";
#define INF 1e17+25
#define pii pair<int,int>
#define vi vector<int>

typedef long long ll;
typedef long double ld;
const ll mod=1e9+7;

ll fpow(ll a, ll b) {

  ll res=1;
  while(b) {
    if(b&1)
      res=(res*a);
    a=(a*a);
    b>>=1;
  }
  return res;
}

ll minm(ll a,ll b){
  if(a<b) return a;
  return b;
}

ll maxm(ll a,ll b){
  if(a>b){ return a;}
  else return b;
}

int PHI(int k){
  // This is Euler's Totient Function.
  //This function returns no. of coprime numbers to N in range [1,2,....,N]

    int ans=k;
    for(int i=2;i*i<=k;++i){
      if(k%i==0){
        ans-=ans/i;
        while(k%i==0){
          k/=i;
        }
      }
    }
     if(k>1) ans-=ans/k;
     return ans;
 

}

int ask(int l,int r){
	cout<<"? "<<l<<' '<<r<<endl;
	int x; cin >> x;
	return x;
}

void solve(){
   int n;
   cin >> n;
   
   int overall=ask(1,n);
   if(overall!=1 && ask(1,overall)==overall){
   	 // 1....overall
   	 
   	 int l=1,r=overall;
   	 while(l!=r-1){
   	 	int mid = (l+r)/2;
   	 	if(ask(mid,overall)==overall){
   	 		l=mid;
   	 	}
   	 	else{
   	 		r=mid;
   	 	}
   	 }
   	 if(ask(l,r)==r)
   	 cout<<"! "<<l<<endl;
   	 else cout<<"! "<<r<<endl;
   }
   else{
   	// overall...n 
   	
   	int l=overall,r=n;
   	 while(l!=r-1){
   	 	int mid = (l+r)/2;
   	 	if(ask(overall,mid)==overall){
   	 		r=mid;
   	 	}
   	 	else{
   	 		l=mid;
   	 	}
   	 }
   	 
   	 	 if(ask(l,r)==r)
   	 cout<<"! "<<l<<endl;
   	 else cout<<"! "<<r<<endl;
   }
 
}


signed main(void)     
{

  fast;

 //  #ifndef ONLINE_JUDGE
    //freopen("input.txt","r",stdin);
  //  freopen("output.txt","w",stdout);
 //   #endif

  int _=1;
 // cin>>_;
 
      for(int test_case=1;test_case<=_;++test_case){
         solve();
      }
   
  
}