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 JAVA):

Solution:
//Implemented By Aman Kotiyal Date:-19-Feb-2021  Time:-1:07:15 am
import java.io.*;
import java.util.*;

public class ques1
{
	public static void main(String[] args)throws Exception{ new ques1().run();} 
	long mod=1000000000+7;
	
	void solve() throws Exception
	{
		int n=ni();
		int l=1,r=n;
		out.println("? "+l+" "+r);
		out.flush();
		int ind=ni();
		int sec = ind;
		int cnt=1;
		if(sec!=1&&sec!=n)
		{
			out.println("? "+l+" "+sec);
			out.flush();
			ind=ni();
			if(ind==sec) // left side
				r=sec;
			else// right side
				l=sec;
			cnt++;
		}
		int ans=1;
		if(sec==n-1)
			ans=n;
		
		while(l<=r)
		{
			int mid=(l+r)/2;					
			if(sec<mid)
			{
				cnt++;
				out.println("? "+sec+" "+mid);
				out.flush();
				ind=ni();
				if(ind==sec)
					{
					r=mid-1;
					ans=mid;
					}
				else
					{
					l=mid+1;
					ans=mid+1;
					}
			}
			else if(sec>mid)
			{
				cnt++;
				out.println("? "+mid+" "+sec);
				out.flush();
				ind=ni();
				if(ind==sec)
					{
					l=mid+1;
					ans=mid;
					}
				else
					{
					r=mid-1;
					ans=mid-1;
					}
			}
			else
				{
				if(l+1==r)
				{
					out.println("? "+l+" "+r);
					out.flush();
					ind=ni();
					if(sec==ind)
					{
						if(sec==l)
							ans=r;
						if(sec==r)
							ans=l;
					}
					break;
				}
				break;
				}
		}
		out.println("! "+ans);
		out.flush();
	}
	
	
	/*FAST INPUT OUTPUT & METHODS BELOW*/
	
	private byte[] buf=new byte[1024];
	private int index;
	private InputStream in;
	private int total;
	private SpaceCharFilter filter;
	PrintWriter out;
	
	int min(int... ar){int min=Integer.MAX_VALUE;for(int i:ar)min=Math.min(min, i);return min;}
	long min(long... ar){long min=Long.MAX_VALUE;for(long i:ar)min=Math.min(min, i);return min;}
	int max(int... ar) {int max=Integer.MIN_VALUE;for(int i:ar)max=Math.max(max, i);return max;}
	long max(long... ar) {long max=Long.MIN_VALUE;for(long i:ar)max=Math.max(max, i);return max;}
	void reverse(int a[]){for(int i=0;i<a.length>>1;i++){int tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
	void reverse(long a[]){for(int i=0;i<a.length>>1;i++){long tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
	String reverse(String s){StringBuilder sb=new StringBuilder(s);sb.reverse();return sb.toString();}
	
	void shuffle(int a[]) {
		ArrayList<Integer> al = new ArrayList<>();
		for(int i=0;i<a.length;i++) 
			al.add(a[i]);
		
		Collections.sort(al);
		for(int i=0;i<a.length;i++) 
			a[i]=al.get(i);
	}
	long lcm(long a,long b)
	{
		return (a*b)/(gcd(a,b));
	}
	
	int gcd(int a, int b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b%a, a); 
	} 
	long gcd(long a, long b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b%a, a); 
	}
	/* for (1/a)%mod = ( a^(mod-2) )%mod  ----> use expo to calc -->(a^(mod-2)) */
	long expo(long p,long q)  /*  (p^q)%mod   */
	{
		long z = 1;
		while (q>0) {
			if (q%2 == 1) {
				z = (z * p)%mod;
			}
			p = (p*p)%mod;
			q >>= 1;
		}
		return z;
	}
	void run()throws Exception
	{
		in=System.in; out = new PrintWriter(System.out);
		solve();
		out.flush();
	}
	private int scan()throws IOException
	{
		if(total<0)
			throw new InputMismatchException();
		if(index>=total)
		{
			index=0;
			total=in.read(buf);
			if(total<=0)
				return -1;
		}
		return buf[index++];
	}
	private int ni() throws IOException 
	{
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = scan();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = scan();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
	private long nl() throws IOException 
	{
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = scan()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = scan();
		}
		
		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = scan();
		}
	}
	private double nd() throws IOException{
		return Double.parseDouble(ns());
	}
	private String ns() throws IOException {
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		StringBuilder res = new StringBuilder();
		do {
			if (Character.isValidCodePoint(c))
				res.appendCodePoint(c);
			c = scan();
		} while (!isSpaceChar(c));
		return res.toString();
	}
	private String nss() throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		return br.readLine();
	}
	private char nc() throws IOException 
	{
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		return (char) c;
	}
	private boolean isWhiteSpace(int n)
	{
		if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
			return true;
		return false;
	}
	private boolean isSpaceChar(int c) {
		if (filter != null)
			return filter.isSpaceChar(c);
		return isWhiteSpace(c);
	}
	private interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}
}