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.StringTokenizer;

public class C {
	public int ask(IO io, int l, int r) throws IOException {
		if (l == r) {
			return r + 1;
		}
		io.println("? " + l + " " + r);
		io.flush();
		return io.nextInt();
	}

    public void run() throws IOException {
		IO io = new IO();
		int n = io.nextInt();
		int d = ask(io, 1, n);
		if (ask(io, 1, d) == d) {
			int l = 1;
			int r = d;
			int m;
			while (r - l > 1) {
				m = (r + l) / 2;
				if (ask(io, m, d) == d) {
					l = m;
				} else {
					r = m;
				}
			}
			io.println("! " + l);
			io.flush();
		} else {
			int l = d;
			int r = n;
			int m;
			while (r - l > 1) {
				m = (r + l) / 2;
				if (ask(io, d, m) == d) {
					r = m;
				} else {
					l = m;
				}
			}
			io.println("! " + r);
			io.flush();
		}
		io.close();
    }

    public static void main(String[] args) throws IOException {
        new C().run();
    }

}

class IO {
	public IO() {
        br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

	public void print(String a) {
		out.print(a);
	}

	public void println(String a) {
		out.println(a);
	}

	public void close() {
		out.close();
	}

    public void flush() {
        out.flush();
    }
}