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 <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <stdlib.h>
#include <stack>
#include <queue>
#define ri register int

inline int read() {
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9') {if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9') {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
	return x*f;
}

void write(int x){
  if(x>9) write(x/10);
  putchar(x%10^48);
}

void Clear(){fflush(stdout);}

int query(int l,int r){
	printf("? %d %d\n",l,r),Clear();
	int k=read();
	return k;
}

int main(){
	int n=read(),l=1,r=n,ans;
	int mex=query(1,n);
	if(mex==1||query(1,mex)!=mex){
		l=mex+1;
		while(l<=r){
//			if(l==r){printf("! %d\n",l),Clear(),exit(0);} 
			int mid=l+r>>1;
			int ls=query(mex,mid);
			if(ls==mex)r=mid-1,ans=mid;
			else l=mid+1;
		}
	}
	else {
		r=mex-1;
		while(l<=r){
//			if(l==r){printf("! %d\n",r),Clear(),exit(0);} 
			int mid=l+r>>1;
			int ls=query(mid,mex);
			if(ls==mex)l=mid+1,ans=mid;
			else r=mid-1;
		}
	}
	printf("! %d\n",ans),Clear();
	return 0;
}