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:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class C2_Rnd703_Div2 
{
	static int n;
	public static void main(String[] args) throws NumberFormatException, IOException
	{
		BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);

		n = Integer.parseInt(scan.readLine());
		
		int ans = solve(n, scan, out);
		out.println("! " + (ans+1));

		out.flush();
	}

	public static int parse(String num)
	{
		return Integer.parseInt(num);
	}
	
	public static int solve(int n, BufferedReader scan, PrintWriter out) throws IOException
	{
		int smax = query(0, n-1, scan, out);
		
		if(smax == 0)
			return binSearchRight(smax, scan, out);
		else if(smax == n-1)
			return binSearchLeft(smax, scan, out);
		
		else if(query(0, smax, scan, out) == smax)
			return binSearchLeft(smax, scan, out);
		else
			return binSearchRight(smax, scan, out);
	}
	
	public static int binSearchLeft(int smax, BufferedReader scan, PrintWriter out) throws IOException 
	{
		int low = 1, high = smax - 1, mid;
		int index = 0;
		
		while(low <= high)
		{
			mid = (low + high) / 2;
			
			// smax is in this range
			if(sendQuery(mid, smax, scan, out) == smax)
			{
				low = mid + 1;
				
				if(index < mid)
					index = mid;
			}
			
			else
			{
				high = mid - 1;
			}
		}
		
		return index;
	}
	
	// need to make sure this doesn't make a query of length 1
	private static int binSearchRight(int smax, BufferedReader scan, PrintWriter out) throws IOException 
	{
		int low = smax + 1, high = n - 2, mid;
		int index = n - 1;
		
		while(low <= high)
		{
			mid = (low + high) / 2;
			
			// smax is in that range
			if(sendQuery(mid, smax, scan, out) == smax)
			{
				high = mid - 1;
				
				if(index > mid)
					index = mid;
			}
			
			else
			{
				low = mid + 1;
			}
		}
		
		return index;
	}

	private static int sendQuery(int a, int b, BufferedReader scan, PrintWriter out) throws IOException 
	{
		if(a > b)
			return query(b, a, scan, out);
		else
			return query(a, b, scan, out);
	}

	public static int query(int l, int r, BufferedReader scan, PrintWriter out) throws IOException
	{
		out.println("? " + (l+1) + " " + (r+1));
		out.flush();
		return parse(scan.readLine()) - 1;
	}
}