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;
using ll=long long;
int main(){
    int n;cin>>n;
    int l=0;int r=n-1;
    bool right=false,left=false;
    cout<<'?'<<" "<<l+1<<" "<<r+1<<"\n";
    cout.flush();
    int sh;cin>>sh;
    if(n==2) cout<<'!'<<" "<<n-sh+1;
    else{
int csh;
    if(sh>1){
    cout<<'?'<<" "<<1<<" "<<sh<<"\n";
    cout.flush();
    cin>>csh;
    if(csh==sh) {r=sh-1;left=true;}
    else {l=sh-1;right=true;}}
    else  {l=sh-1;right=true;}
    bool ans=true;
    while(ans){
        if(left){
            int m=(l+r)/2;
            if(l==m) {ans=false;break;}
            cout<<'?'<<" "<<m+1<<" "<<sh<<"\n";
            cout.flush();
            cin>>csh;
            if(csh==sh) {l=m;}
            else r=m;

        }
        else{
            int m=(l+r)/2;
            if((l+r)%2!=0) m++;
            if(r==m) {ans=false;l=m;break;}
            cout<<'?'<<" "<<sh<<" "<<m+1<<"\n";
            cout.flush();
            cin>>csh;
            if(csh==sh) {r=m;}
            else l=m;

        }

    }
    cout<<'!'<<" "<<l+1;}}