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.util.*;
import java.io.*;

public class C1486 {
	public static void main(String[] args) throws Exception {
		int n = i();
		out.println("? 1 " + n);
		out.flush();
		int sec = i();
		int sectemp = 0;
		
		if (sec!=1) {
			out.println("? 1 " + sec);
			out.flush();
			sectemp = i();
		}
		
		if (sec==sectemp && sectemp!=0) {
			int front = 1; int end = sec-1;
			while (end!=front) {
				int mid = (int) Math.ceil((front+end)/2.0);
				out.println("? " + mid + " " + sec);
				out.flush();
				
				int temp = i();
				if (temp==sec) front = mid;
				else end = mid-1;
			}
			
			out.println("! " + front);
		}
		
		else {
			int front = sec+1; int end = n;
			while (end!=front) {
				int mid = (front+end)/2;
				out.println("? " + sec + " " + mid);
				out.flush();
				
				int temp = i();
				if (temp==sec) end = mid;
				else front = mid+1;
			}
			
			out.println("! " + front);
		}
		
		in.close();
		out.close();
	}
	
	static BufferedReader in;
	static StringTokenizer st = new StringTokenizer("");
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	// static PrintWriter out;
	/*
	 * static { try { in = Files.newBufferedReader(Paths.get("haybales.in")); out =
	 * new PrintWriter(new BufferedWriter(new FileWriter("haybales.out"))); } catch
	 * (Exception e) { in = new BufferedReader(new InputStreamReader(System.in)); }
	 * }
	 */
	static {
		in = new BufferedReader(new InputStreamReader(System.in));
	}

	static void check() throws Exception {
		while (!st.hasMoreTokens())
			st = new StringTokenizer(in.readLine());
	}

	static String s() throws Exception {
		check();
		return st.nextToken();
	}

	static int i() throws Exception {
		return Integer.parseInt(s());
	}

	static long l() throws Exception {
		return Long.parseLong(s());
	}

	static double d() throws Exception {
		return Double.parseDouble(s());
	}
}