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 INF 2000000000
typedef long long int ll;
typedef pair<ll,ll> ii;
typedef pair<ii,ll> iii;
typedef vector<ll> vi;
typedef vector<ii> vii;
#define getbit(n,i) (((n)&(1LL<<(i)))!=0)
#define setbit0(n,i) ((n)&(~(1LL<<(i))))
#define setbit1(n,i) ((n)|(1LL<<(i)))
#define lastone(n) ((n)&(-(n)))
#define read freopen("debug//in.txt","r",stdin)
#define write freopen("debug//out.txt","w",stdout)
#define DBG(a) cerr<<#a<<" ->->->-> "<<a<<"\n"
#define fi first
#define se second
#define PI (acos(-1))
#define fastread ios::sync_with_stdio(false);cin.tie(NULL)
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define mod (1000000000)
#define asz 100005
template<class T,class V> ostream& operator<<(ostream &s,pair<T,V> a)
{
    s<<a.fi<<' '<<a.se;
    return s;
}

const int N = 1e5+5;
int ask(int l,int r){
    if(l>=r)return -1;
    cout<<"? "<<l<<' '<<r<<endl;
    int x;
    cin>>x;
    return x;
}
int main()
{
    fastread;
#ifdef FILE_IO2
    read;
    write;
#endif

    int T=1;
//    cin>>T;
    for(int qq=1; qq<=T; qq++)
    {
        int n;
        cin>>n;
        int a = ask(1,n);
        if(ask(a,n)==a){
            int l = a+1,r = n;
            while(l<r){
                int mid = (l+r)/2;
                if(ask(a,mid)==a)r = mid;
                else l = mid+1;
            }
            cout<<"! "<<l<<endl;
        }
        else{
            int l = 1,r = a-1;
            while(l<r){
                int mid = (l+r+1)/2;
                if(ask(mid,a)==a)l = mid;
                else r = mid-1;
            }
            cout<<"! "<<l<<endl;
        }
    }
}