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.*;
import java.util.*;

public class MainC {
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }
        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public boolean hasNext() {
            try {
                String string = reader.readLine();
                if (string == null) {
                    return false;
                }
                tokenizer = new StringTokenizer(string);
                return tokenizer.hasMoreTokens();
            } catch (IOException e) {
                return false;
            }
        }
    }
    static InputReader in = new InputReader(System.in);
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    public static void main(String[] args) {
        solve();
    }
    static int answer;
    static void solve() {
        int n = in.nextInt();
        dg1(1, n);
    }
    static void dg1 (int start, int end) {
        int secIndex = query(start, end);
        if (start + 1 == end) {
            answer = secIndex == start ? end : start;
            a(answer);
            return;
        }
        boolean flag = false;
        if (secIndex == 1) {
            flag = true;
        }
        int temp = 0;
        if (!flag) temp = query(start, secIndex);
        int low, high, mid;
        if (flag || temp != secIndex) {
            low = secIndex; high = end;
            while(low + 1 < high) {
                mid = ((low + high)/2);
                if (secIndex == query(secIndex, mid)) high = mid;
                else low = mid;
            }
            a(high);
        } else {
            low = 1; high = secIndex;
            while(low + 1 < high) {
                mid = ((low + high)/2);
                if (secIndex == query(mid, secIndex)) low = mid;
                else high = mid;
            }
            a(low);
            
        }
    }
    static int query (int s, int e) {
        System.out.println("? " + s + " " + e);
        System.out.flush();
        return in.nextInt();
    }
    static void a (int index) {
        System.out.println("! " + index);
    }
}