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.util.*;
import java.io.*;

public class Main {
    static FastScanner sc = new FastScanner(System.in);
    static PrintWriter pw = new PrintWriter(System.out);
    static StringBuilder sb = new StringBuilder();

    public static void main(String[] args) throws Exception {
        int n = sc.nextInt();
        System.out.println("? 1 " + n);
        int center = sc.nextInt();
        if (center == 1) {
            int left = 1;
            int right = n;
            while (right - left > 1) {
                int mid = (left + right) / 2;
                System.out.println("? " + center + " " + mid);
                int now = sc.nextInt();
                if (center == now) {
                    right = mid;
                } else {
                    left = mid;
                }
            }
            System.out.println("! " + right);
        } else if (center == n) {
            int left = 1;
            int right = n;
            while (right - left > 1) {
                int mid = (left + right) / 2;
                System.out.println("? " + mid + " " + center);
                int now = sc.nextInt();
                if (center == now) {
                    left = mid;
                } else {
                    right = mid;
                }
            }
            System.out.println("! " + left);
        } else {
            System.out.println("? 1 " + center);
            int now = sc.nextInt();
            if (center == now) {
                int left = 1;
                int right = center;
                while (right - left > 1) {
                    int mid = (left + right) / 2;
                    System.out.println("? " + mid + " " + center);
                    now = sc.nextInt();
                    if (center == now) {
                        left = mid;
                    } else {
                        right = mid;
                    }
                }
                System.out.println("! " + left);
            } else {
                int left = center;
                int right = n;
                while (right - left > 1) {
                    int mid = (left + right) / 2;
                    System.out.println("? " + center + " " + mid);
                    now = sc.nextInt();
                    if (center == now) {
                        right = mid;
                    } else {
                        left = mid;
                    }
                }
                System.out.println("! " + right);
            }
        }

    }

    static class GeekInteger {
        public static void save_sort(int[] array) {
            shuffle(array);
            Arrays.sort(array);
        }

        public static int[] shuffle(int[] array) {
            int n = array.length;
            Random random = new Random();
            for (int i = 0, j; i < n; i++) {
                j = i + random.nextInt(n - i);
                int randomElement = array[j];
                array[j] = array[i];
                array[i] = randomElement;
            }
            return array;
        }

        public static void save_sort(long[] array) {
            shuffle(array);
            Arrays.sort(array);
        }

        public static long[] shuffle(long[] array) {
            int n = array.length;
            Random random = new Random();
            for (int i = 0, j; i < n; i++) {
                j = i + random.nextInt(n - i);
                long randomElement = array[j];
                array[j] = array[i];
                array[i] = randomElement;
            }
            return array;
        }

    }
}

class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner(InputStream in) {
        reader = new BufferedReader(new InputStreamReader(in));
        tokenizer = null;
    }

    public String next() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                tokenizer = new StringTokenizer(reader.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken();
    }

    public String nextLine() {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return tokenizer.nextToken("\n");
    }

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

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

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

    public String[] nextArray(int n) {
        String[] a = new String[n];
        for (int i = 0; i < n; i++)
            a[i] = next();
        return a;
    }

    public int[] nextIntArray(int n) {
        int[] a = new int[n];
        for (int i = 0; i < n; i++)
            a[i] = nextInt();
        return a;
    }

    public long[] nextLongArray(int n) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++)
            a[i] = nextLong();
        return a;
    }
}