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>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define fore(i,a,b) for(i=(a);i<=(int)(b);i++)
#define ford(i,a,b) for(i=(a);i>=(int)(b);i--)
#define Cerr if(!debug_cerr) {} else cerr
#define loger(...) Cerr << '>' << #__VA_ARGS__  << ": "; dbg_out(__VA_ARGS__ );
using namespace std; bool debug_cerr=1;
void dbg_out() { Cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { Cerr<<H<<","; dbg_out(T...); }
template<class A,class B> ostream& operator<<(ostream &os,const pair<A, B> &p){return os<<'('<<p.first<<","<<p.second<<')';}
template <typename T> ostream& operator<<(ostream& stream,const vector<T>& obj){
    stream<<"[";for(auto x:obj)stream<<x<<",";stream<<"]"<<((obj.size()>3)?"\n":""); return stream; }
template <typename T,size_t N> ostream& operator<<(ostream& stream,const array<T,N>& obj){
    stream<<"[";for(auto x:obj)stream<<x<<",";stream<<"]"<<((N>3)?"\n":"");return stream;}
template <typename T> void printarr(T arr[],int n){Cerr<<"Array ";for(int i=0;i<n;i++){Cerr<<arr[i]<<" ";}Cerr<<endl;}
template <typename T> void printarr(T arr){Cerr<<">>";for(auto it=arr.begin();it!=arr.end();it++){Cerr<<*it<<" ";}Cerr<<endl;}
typedef long long ll;typedef unsigned long long ull;
const ll mod=1000000007;
ll sec;
ll query(ll l,ll r){
    cout<<"? "<<l<<' '<<r<<endl;
    ll ind;
    cin>>ind;
    return ind;
}
void red(ll& l,ll& r){
    if(r-l<=0)return;
    if(r<sec){
        ll mid=(l+r)/2;
        ll ind=query(mid+1,sec);
        if(ind==sec){
            l=mid+1;
        }else r=mid;
    }else{
        ll mid=(l+r)/2;
        ll ind=query(sec,mid);
        if(ind==sec){
            r=mid;
        }else l= mid+1;
    }
}

int main()
{
    //ios_base::sync_with_stdio(false);cin.tie(NULL);
    ll test,ti,i,j,k,n,m,q;
        cin>>n;
        sec=query(1,n);
        ll l,r;
        if(sec==1){
            l=sec+1;
            r=n;
        }else if(sec==n || query(1,sec)==sec){
            l=1;
            r=sec-1;
        }else{
            l=sec+1;
            r=n;
        }
        while(l<r){
            red(l,r);
        }
        cout<<"! "<<l<<endl;


    return 0;
}