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:
//#pragma GCC optimize(4)
#include<bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define REP(u) for(int i=p[u];i!=-1;i=e[i].nxt)
#define PII pair<int,int>
#define ll long long
#define pb push_back
#define PLL pair<ll,ll>
#define fi first
#define se second
#define sz(x) (int)(x.size())
#define rd (rand()<<16^rand())
#define db double
#define gc getchar()//(_p1==_p2&&(_p2=(_p1=_buf)+fread(_buf,1,100000,stdin),_p1==_p2)?EOF:*_p1++)
using namespace std;
char _buf[100000],*_p1=_buf,*_p2=_buf;
inline int gi()
{
	int x=0,f=1;
	char ch=gc;
	while(ch<'0'||ch>'9')
	{
		if(ch=='-')f=-1;
		ch=gc;
	}
	while(ch>='0'&&ch<='9')
	{
		x=(x<<3)+(x<<1)+(ch^48);
		ch=gc;
	}
	return (f==1)?x:-x;
}
inline int max(int a,int b){return a>b?a:b;}
inline int min(int a,int b){return a<b?a:b;}
const int maxn=1e5+5;
int a[maxn];
inline int query(int l,int r)
{
	printf("? %d %d\n",l,r);
//	int maxv=0,max2=0,pp=0,pp2=0;
//	FOR(i,l,r)
//	{
//		if(a[i]>maxv)maxv=a[i],pp=pp2,pp2=i;
//		else if(a[i]>max2)max2=a[i],pp=i;
//	}
//	cout<<pp<<endl;
//	return pp;
	fflush(stdout);
	int ret=gi();
	return ret;
}
inline void output(int x)
{
	printf("! %d\n",x);
	fflush(stdout);
}
int n;
inline void init()
{

}
inline void input()
{
	n=gi();
}
inline void solve()
{
	int now=query(1,n);
	int l=1,r=n;
	if(now!=n&&now!=1)
	{
		int now2=query(1,now);
		if(now2==now)r=now-1;
		else l=now+1;
	}
	else if(now==1)l=2;
	else if(now==n)r=n-1;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(l==r)
		{
			output(l);
			return ;
		}
		if(l==r-1)
		{
			now=query(l,r);
			if(now==l)output(r);
			else output(l);
			return ;
		}
		if(now==mid)mid++;
		if(now<mid)
		{
			int now1=query(now,mid);
			if(now1==now)r=mid;
			else l=mid+1;
		}
		else
		{
			int now1=query(mid,now);
			if(now1==now)l=mid;
			else r=mid-1;
		}
	}
}
int main()
{
	int T=1;
//	T=gi();
	while(T--)
	{
		init();
		input();
		solve();
	}
	return 0;
}