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:
// i_m_arin

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef vector<long long> vll;
typedef vector<string> vs;
typedef priority_queue<ll> pqll;
typedef pair<ll,ll> pll;
typedef vector<pll> vpll;
typedef map<ll,ll> mpll;
typedef set<ll> stll;

#define f(i, a, n)   for(ll i=a; i<(ll)n; i++)
#define f2(i, a, b) for(ll i=a; i<=(ll)b; i++)
#define pb       push_back
#define ff       first
#define ss       second
#define pf      push_front
#define mp      make_pair
#define all(a)    (a).begin(),(a).end()
#define decimal(x)  fixed<<setprecision(x)
#define fsort(a)  sort(all(a))
#define rsort(a)  sort(all(a),greater<>())
#define fast_io    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define mem0(a)    memset(a,0,sizeof(a))
#define mem1(a)    memset(a,-1,sizeof(a))
#define lcm(a,b)  (a*b)/__gcd(a,b)
#define ub(v,x)    upper_bound(all(v),x)-v.begin()
#define lb(v,x)    lower_bound(all(v),x)-v.begin()

const ld PI  = acos(-1);
const ll MOD = 1e9+7;
const ll INF = LLONG_MAX;
const ll SIZE= 100000;

ll func(ll start,ll end){
	cout<<"? "<<start<<" "<<end<<"\n";
	cout.flush();
	ll x;
	cin>>x;
	return x;
}



void solve(){

ll n;
cin>>n;

if(n==2){
	ll z=func(1,2);
	if(z==1){
		cout<<"! "<<2<<endl;
	}
	else{
		cout<<"! "<<1<<endl;
	}
	return;
}

ll x=func(1,n);
ll y=-1;
if(x!=1)
	y=func(1,x);
if(x==1 || y!=x){
	ll l=x,r=n;
	while(l!=r){
		if(r-l==1){
			ll z=func(l,r);
			if(z==l) break;
			else{
				r=l;
				break;
			}
		}
		ll mid=ceil((l+r)*1.0/2);
		ll some=func(x,mid);
		if(some==x){
			r=mid;
		}
		else l=mid;
	}
	cout<<"! "<<r<<endl;
}
else{
	ll l=1;
	ll r=x;
	while(l!=r){
		if(r-l==1){
			ll z=func(l,r);
			if(z==l) break;
			else{
				r=l;
				break;
			}
		}
		ll mid=(l+r)/2;
		ll some=func(mid,x);
		if(some==x){
			l=mid;
		}
		else r=mid;
	}
	cout<<"! "<<r<<endl;
}


}


int main(){
  fast_io;
  int t=1;
  //cin>>t;
  while(t--){
    solve();
  }
  return 0;
}