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;
import java.util.*;

public class GuessingTheGreatestHard {
	static FastScanner fs = new FastScanner();
	static int query(int left, int right) {
		System.out.println("? " + left + " " + right);
		return fs.nextInt();
	}

	public static void main(String[] args) {
		PrintWriter out = new PrintWriter(System.out);
		int n = fs.nextInt();
		int secondMax = query(1, n);
		int left = 1, right = n;
		if(secondMax == 1 || query(1, secondMax) != secondMax) {
			left = secondMax;
			right = n;
			while(left + 1 < right) {
				int mid = (left + right) / 2;
				if(query(secondMax, mid) == secondMax)
					right = mid;
				else
					left = mid;
			}
		} else {
			left = 1;
			right = secondMax;
			while(left + 1 < right) {
				int mid = (left + right) / 2;
				if(query(mid, secondMax) == secondMax)
					left = mid;
				else
					right = mid;
			}
		}
		int myAns = query(left, right) == left ? right : left;
		System.out.println("! " + myAns);
	}
	static void ruffleSort(int[] a) {
		int n = a.length;
		Random r = new Random();
		for(int i = 0; i < a.length; ++i) {
			int oi = r.nextInt(n), t = a[i];
			a[i] = a[oi];
			a[oi] = t;
		}
		Arrays.sort(a);
	}
	static void sort(int[] a)   {
		ArrayList<Integer> l = new ArrayList<>();
		for(int i : a) l.add(i);
		Collections.sort(l);
		for(int i = 0; i < a.length; ++i) a[i] = l.get(i);
	}
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next()   {
			while(!st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a = new int[n];
			for(int i = 0; i < n; ++i)
				a[i] = nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}