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 pb push
#define speed ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector <ll> vll;
typedef pair<ll,ll> pll;
typedef vector <pair<ll,ll>> vpll;
typedef vector <vll> vvll;
ll n;

int get(int x,int y)
{
    int val;
    cout<<"? "<<x<<" "<<y<<endl;
    cin>>val;
    return val;
}

int main()
{
    speed;
    cin>>n;
    ll l,h,m,midx,i;
    midx=get(1,n);
    
    if(midx>1 && get(1,midx)==midx)
    {
        l=1;
        h=midx-1;
        
        while(l<h)
        {
            m=(l+h+1)/2;
            i=get(m,midx);
            if(i==midx)
                l=m;
            else 
                h=m-1;
        }
        cout<<"! "<<l<<endl;
    }
    else
    {
        l=midx+1;
        h=n;
        
        while(l<h)
        {
            m=(l+h)/2;
            i=get(midx,m);
            if(i!=midx)
                l=m+1;
            else
                h=m;
        }
        cout<<"! "<<l<<endl;
    }
    return 0;
}