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;
//#include "testlib.h"
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define int long long
#define ll long long
#define M 1000000007
#define MM 998244353
#define inputarr(a,n) for(int i=0;i<n;++i) cin>>a[i]
#define GCD(m,n) __gcd(m,n)
#define LCM(m,n) m*(n/GCD(m,n))
#define mii  map<ll ,ll >
#define msi  map<string,ll >
#define rep(a,b)    for(ll i=a;i<b;i++)
#define rep0(n)    for(ll i=0;i<n;i++)
#define repi(i,a,b) for(ll i=a;i<b;i++)
#define pb push_back
#define vi vector<ll>
#define vs vector<string>
#define ppb pop_back
#define endl '\n'
#define asdf ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define r0 return 0;
#define FORD(i, a, b) for (int i = (int) (a); i >= (int) (b); --i)
#define inputoutput freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define Set(a, s) (a, s, sizeof (a))
#define FOR repi
#define vii vector<pii>
#define pii pair<int,int>
#define REVERSE(v) reverse(all(v))
#define trav(a, x) for(auto& a : x)
#define display(x) trav(a,x) cout<<a<<" ";cout<<endl
#define debug cerr<<"bhau"<<endl
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');std::cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
template<typename T, typename U> static inline void amin(T &x, U y) 
{ 
    if (y < x) 
        x = y; 
}
template<typename T, typename U> static inline void amax(T &x, U y) 
{ 
    if (x < y) 
        x = y; 
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

ll max(ll a, ll b) { return (a > b)? a : b;}
int min(int a, int b) { return (a < b)? a : b;} 



int query(int l,int r){
	cout<<"? "<<l<<" "<<r<<endl;
	int x;
	cin>>x;
	return x;
}
void answer(int x){
	cout<<"! "<<x<<endl;
}
int solve(){

	int n;
	cin>>n;
	int l = 1, r = n;

	int ind = query(l,r);

	if(ind!=1 && query(1,ind)==ind){
		l = 1, r= ind - 1;
		int ans = 1;
		while(l<=r){
			int mid = l + (r-l)/2;
			int z = query(mid,ind);
			// trace(mid,ind,z,ans,l,r);
			if(z==ind){
				ans = mid;
				l = mid + 1;
			}
			else{
				r = mid-1;
			}
		}
		answer(ans);
		return 0;
	}
	l = ind+1, r= n;
		int ans = n;
		while(l<=r){
			int mid = l + (r-l)/2;
			int z = query(ind,mid);
			if(z==ind){
				ans = mid;
				r = mid - 1;
			}
			else{
				l = mid + 1;
			}

		}
		answer(ans);



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