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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Stack;
import java.util.StringTokenizer;
import static java.lang.Math.*;

public class C {

	static FastScanner sc;
	static PrintWriter pw;

	public static void main(String[] args) {
		sc = new FastScanner();		
		pw = new PrintWriter(System.out);
		int T = 1;
		while (T-- > 0) {
			solve();
		}
		pw.close();
	}

	static HashMap<Integer, HashMap<Integer, Integer>> rangeMap = new HashMap<>();
	
	static void solve() {
		int n = sc.nextInt();
		int low = 1, high = n;
		rangeMap.clear();		
		if(high - low > 1) {
			int secondMax = getQuery(low, high);
			if(secondMax != high) {
				int secondMax2 = getQuery(secondMax, high);				
				if(secondMax2 == secondMax) {
					low = secondMax + 1;					
				} else {
					high = secondMax - 1;
				}
			} else {
				high = secondMax - 1;
			}
			while(high - low > 1) {			
				int mid = (low + high) / 2;
				
				int ind = getQuery(min(secondMax, mid), max(secondMax, mid));
				
				if(ind == secondMax) {
					if(mid > secondMax) {					
						high = mid;
					} else {					
						low = mid;
					}
				} else {
					if(mid > secondMax) {
						low = mid + 1;
					} else {
						high = mid - 1;
					}
				}
				
			}
		}
		int ans = -1;
		if(high == low + 1) {
			int notAnswer = getQuery(low, high);
			ans = notAnswer == low ? high : low;
		}else {
			ans = low;
		}
		System.out.println("! "+ans);
		
	}
	
	static int getQuery(int low, int high) {
		if(rangeMap.containsKey(low)) {
			if(rangeMap.get(low).containsKey(high)) {
				return rangeMap.get(low).get(high);
			}
		}
				
		System.out.println("? "+low+" "+high);
		int num = sc.nextInt();
		
		HashMap<Integer, Integer> map = rangeMap.getOrDefault(low, new HashMap<>());
		map.put(high, num);
		rangeMap.put(low, map);
		
		return num;
	}
	
	static int getMove(int low, int ind, int high) {
		if(ind <= low) {
			return 1;
		}		
		int left = getQuery(low, ind);
		if(high <= ind) {
			return -1;
		}	
		int right = getQuery(ind, high);
		if(left == ind) {
			return -1;
		}
		return 1;
	}
	
	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());
		}
	}

}