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:
// Problem: C1. Guessing the Greatest (easy version)
// Contest: Codeforces - Codeforces Round #703 (Div. 2)
// URL: https://codeforces.com/contest/1486/problem/C1
// Memory Limit: 256 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
#define rep(i,x,y) for(auto i=(x);i<=(y);++i)
#define dep(i,x,y) for(auto i=(x);i>=(y);--i)
#define fr first
#define sc second
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,ll>pil;
typedef pair<int,int>pii;
typedef pair<ll,ll>pll;
const int N=1e6+10;
const int inf=1e9;
const int mo=998244353;
ll a[N];
int query(int l,int r){
    cout<<"? "<<l<<" "<<r<<endl;
    cout.flush();
    int x; cin>>x;
    return x;
}
int main(){
    //freopen("D:\\in.txt","r",stdin);
    //freopen("D:\\out.txt","w",stdout);
    int n; cin>>n;
	int p=query(1,n);
	int l=1,r=n,mid;
	if(p!=1&&p==query(1,p)){
		r=p;
		while(r-l>1){
			mid=l+r>>1;
			if(query(mid,p)==p) l=mid;
			else r=mid;  
		}
	}
	else{
		l=p;
		while(r-l>1){
			mid=l+r>>1;
			if(query(p,mid)==p) r=mid;
			else l=mid;  
		}
	}
	p=query(l,r);
	if(p==l) p=r;
	else p=l;
	cout<<"! "<<p<<endl;
    cout.flush();
    return 0;
}