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.util.StringTokenizer;

public class C {
    public static void main(String[] args) {
        FastScanner fs = new FastScanner();
        int t = 1;//fs.nextInt();
        while (t-- > 0) {
            int n = fs.nextInt();
            int overall = query( 1, n, fs );
            boolean left = false;
            if (overall != 1) {
                if (query( 1, overall, fs ) == overall) left = true;
            }
            if (left) {
                int low = 1, high = overall - 1;
                while (low <= high) {
                    int mid = (low + high) >> 1;
                    if (query( mid, overall, fs ) == overall) low = mid + 1;
                    else high = mid - 1;
                }
                System.out.println( "! " + high );
                System.out.flush();

            } else {
                int low = overall + 1, high = n;
                while (low <= high) {
                    int mid = (low + high) >> 1;
                    if (query( overall, mid, fs ) == overall) high = mid - 1;
                    else low = mid + 1;
                }
                System.out.println( "! " + low );
                System.out.flush();
            }
        }
    }

    static int query(int low, int high, FastScanner fs) {
        System.out.println( "? " + low + " " + high );
        System.out.flush();
        return fs.nextInt();
    }

    private static class FastScanner {
        BufferedReader br = new BufferedReader( new InputStreamReader( System.in ) );
        StringTokenizer st = new StringTokenizer( "" );

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

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

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

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