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.*;

/*
polyakoff
*/

public class Main {

    static FastReader in;
    static PrintWriter out;
    static Random rand = new Random();
    static final int oo = (int) 1e9 + 10;
    static final long OO = (long) 2e18 + 10;
    static final int MOD = (int) 1e9 + 7;

    static int ask(int l, int r) {
        out.printf("? %d %d\n", l, r);
        out.flush();
        return in.nextInt();
    }

    static void solve() {
        int n = in.nextInt();

        int l = 1, r = n;
        int second = ask(l, r);
        while (true) {
            if (r - l + 1 == 2) {
                int first = second == l ? l + 1 : l;
                out.printf("! %d\n", first);
                out.flush();
                return;
            } else if (r - l + 1 == 3) {
                int pos = ask(l, l + 1);
                int first;
                if (second != r)
                    first = pos == second ? (second == l ? l + 1 : l) : r;
                else
                    first = pos == l ? l + 1 : l;
                out.printf("! %d\n", first);
                out.flush();
                return;
            }
            int mid = (l + r) / 2;
            if (second <= mid) {
                int pos = ask(l, mid);
                if (pos == second)
                    r = mid;
                else {
                    int l2 = mid + 1, r2 = r;
                    while (l2 < r2) {
                        int mid2 = (l2 + r2) / 2;
                        if (ask(second, mid2) == second)
                            r2 = mid2;
                        else
                            l2 = mid2 + 1;
                    }
                    out.printf("! %d\n", l2);
                    out.flush();
                    return;
                }
            } else {
                int pos = ask(mid + 1, r);
                if (pos == second)
                    l = mid + 1;
                else {
                    int l2 = l, r2 = mid;
                    while (l2 < r2) {
                        int mid2 = (l2 + r2) / 2;
                        if (ask(mid2 + 1, second) == second)
                            l2 = mid2 + 1;
                        else
                            r2 = mid2;
                    }
                    out.printf("! %d\n", l2);
                    out.flush();
                    return;
                }
            }
        }

    }

    public static void main(String[] args) {
        in = new FastReader();
        out = new PrintWriter(System.out);

        int T = 1;
//        T = in.nextInt();
        while (T-- > 0)
            solve();

        out.flush();
        out.close();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        FastReader() {
            this(System.in);
        }
        FastReader(String file) throws FileNotFoundException {
            this(new FileInputStream(file));
        }
        FastReader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String next() {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(nextLine());
            }
            return st.nextToken();
        }
        String nextLine() {
            String line;
            try {
                line = br.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return line;
        }
    }
}