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.*;
import java.util.function.BiFunction;

public class C {
    static FastReader reader = new FastReader();
    static OutputWriter out = new OutputWriter(System.out);


    static int ask(int l, int r) {
        out.println("? " + l + " " + r);out.flush();
        return reader.nextInt();
    }

    static int solve(int l, int r) {
        if(l == r) {
            return l;
        }
        if(l + 1 == r) {
            return ask(l, r) == l ? r : l;
        }
        int t1 = ask(l, r);
        if(t1 != r && ask(t1, r) == t1) {
            //we are at left most point
            l = t1 + 1;
            while(l <= r) {
                int m = (l + r) / 2;
                int t2 = ask(t1, m);
                if(t2 == t1) {
                    r = m - 1;
                } else {
                    l = m + 1;
                }
            }//[1,2,4,3, 5]
            return l;
        } else {
            r = t1 - 1;
            while(l <= r) {
                int m = (l + r) / 2;
                int t2 = ask(m, t1);
                if(t2 == t1) {
                    l = m + 1;
                } else {
                    r = m - 1;
                }
            }
            return r;
            //we are at right most point
        }


    }

    public static void main(String[] args) {
        try {
            int n = reader.nextInt();
            int ans = solve(1, n);
            out.println("! " + ans);

        } finally {
            out.flush();
        }
    }


    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

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

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void print(char[] array) {
            writer.print(array);
        }

        public void print(Object... objects) {
            for (int i = 0; i < objects.length; i++) {
                if (i != 0) {
                    writer.print(' ');
                }
                writer.print(objects[i]);
            }
        }

        public void print(int[] array) {
            for (int i = 0; i < array.length; i++) {
                if (i != 0) {
                    writer.print(' ');
                }
                writer.print(array[i]);
            }
        }

        public void print(double[] array) {
            for (int i = 0; i < array.length; i++) {
                if (i != 0) {
                    writer.print(' ');
                }
                writer.print(array[i]);
            }
        }

        public void print(long[] array) {
            for (int i = 0; i < array.length; i++) {
                if (i != 0) {
                    writer.print(' ');
                }
                writer.print(array[i]);
            }
        }

        public void println(int[] array) {
            print(array);
            writer.println();
        }

        public void println(double[] array) {
            print(array);
            writer.println();
        }

        public void println(long[] array) {
            print(array);
            writer.println();
        }

        public void println() {
            writer.println();
        }

        public void println(Object... objects) {
            print(objects);
            writer.println();
        }

        public void print(char i) {
            writer.print(i);
        }

        public void println(char i) {
            writer.println(i);
        }

        public void println(char[] array) {
            writer.println(array);
        }

        public void printf(String format, Object... objects) {
            writer.printf(format, objects);
        }

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

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

        public void print(long i) {
            writer.print(i);
        }

        public void println(long i) {
            writer.println(i);
        }

        public void print(int i) {
            writer.print(i);
        }

        public void println(int i) {
            writer.println(i);
        }

        public void separateLines(int[] array) {
            for (int i : array) {
                println(i);
            }
        }
    }
}