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.Scanner;
 
public class D {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int mid = ask(1, n, sc);
		if (mid < n && ask(mid, n, sc) == mid) {
			solve(mid, n, false, sc, mid);
		} else {
			solve(1, mid, true, sc, mid);
		}
	}
	
	public static void solve(int a, int b, boolean isRight, Scanner sc, int mid) {
		int l = a, r = b;
		if (isRight) r--;
		else l++;
		int ans = 0;
		while (l <= r) {
			int m = (l+r)/2;
//			System.out.println("l, m, r: "+l+","+m+","+r);
			if (isRight) {
				int res = ask(m, b, sc);
//				System.out.println("res: "+res);
				if (res == mid) {					
					ans = m;
					l = m+1;
				} else 
					r = m-1;
			} else {
				int res = ask(a, m, sc);
//				System.out.println("res: "+res);
				if (res == mid) {
					ans = m;
					r = m-1;
				} else
					l = m+1;
			}
		}
		System.out.println("! "+ans);
	}
	
	public static int ask(int l, int r, Scanner sc) {
		System.out.println("? "+l+" "+r);
		System.out.flush();
		return sc.nextInt();
	}
}