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 IO  ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std; 
 
const int inf=0x3f3f3f3f;
typedef long long ll; 
const int N = 1e5+7;
const ll mod = 1e9+7;       

int main(){ 
    IO;
    int t=1;
    //cin>>t;
    while(t--){  
        int n;
        cin>>n;
        int l=1,r;
        int x;
        cout<<"? 1 "<<n<<endl;
        fflush(stdout);
        cin>>x;
        int k,ans;
        map<int,int>mp;
        if(x==1||x==n){
            if(x==1){
                l=2;
                r=n;
                while(l<r){ 
                    int mid=(l+r)>>1;
                    while(mp[mid])mid--;
                    cout<<"? "<<x<<" "<<mid<<endl;
                    fflush(stdout);
                    cin>>k;
                    if(k==x)r=mid,mp[mid]=1;
                    else l=mid+1;
                }

            }
            else{
                l=1;
                r=n-1;
                while(l<r){ 
                    int mid=(l+r)>>1;
                    while(mp[mid])mid++;
                    cout<<"? "<<mid<<" "<<x<<endl;
                    fflush(stdout);
                    cin>>k;
                    if(k==x)l=mid,mp[mid]=1;
                    else r=mid-1;
                }
            }
        }else{
            cout<<"? 1 "<<x<<endl;
            fflush(stdout);
            cin>>k;
            if(k==x){
                mp[1]=1;
                l=1;
                r=x-1;
                while(l<r){  
                    int mid=(l+r)>>1;
                    while(mp[mid])mid++;
                    cout<<"? "<<mid<<" "<<x<<endl;
                    fflush(stdout);
                    cin>>k;
                    if(k==x)l=mid,mp[mid]=1;
                    else r=mid-1;
                }
            }
            else {
                l=x+1;
                r=n;
                mp[n]=1;
                while(l<r){ 
                    int mid=(l+r)>>1;
                    while(mp[mid])mid--;
                    cout<<"? "<<x<<" "<<mid<<endl;
                    fflush(stdout);
                    cin>>k;
                    if(k==x)r=mid,mp[mid]=1;
                    else l=mid+1;
                }
            } 
        }
        cout<<"! "<<l<<endl;
    }
    return 0;
}