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 _1486_C2 {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(in.readLine());
		int sec = query(1, n, in);
		int lbound = sec + 1;
		int rbound = n;
		if(sec > 1 && query(1, sec, in) == sec) {
			lbound = 1;
			rbound = sec;
		}
		int res = 0;
		while(lbound < rbound) {
			if(rbound == lbound + 1) {
				int sec2 = query(lbound, rbound, in);
				if(sec2 == lbound) res = rbound;
				else res = lbound;
				break;
			}
			int avg = (lbound + rbound) / 2;
			if(avg < sec) {
				int sec2 = query(avg, sec, in);
				if(sec2 == sec) {
					lbound = avg;
				}else {
					rbound = avg - 1;
				}
			}else {
				int sec2 = query(sec, avg, in);
				if(sec2 == sec) {
					rbound = avg;
				}else {
					lbound = avg + 1;
				}
			}	
		}
		if(res == 0) {
			System.out.println("! " + lbound);
		}else {
			System.out.println("! " + res);
		}
		in.close();
	}
	static int query(int l, int r, BufferedReader in) throws IOException {
		System.out.println("? " + l + " " + r);
		System.out.flush();
		return Integer.parseInt(in.readLine());
	}
}