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;

public class C2 {

    public static boolean check(int l, int r, int v, BufferedReader in) throws IOException {
        System.out.printf("? %d %d\n", l, r);
        System.out.flush();
        int ans = Integer.parseInt(in.readLine());
        return ans == v;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(in.readLine());

        System.out.printf("? %d %d\n", 1, n);
        System.out.flush();

        int pos = Integer.parseInt(in.readLine());
        int l, r;
        if(pos > 1 && check(1, pos, pos, in)){
            l = 1;
            r = pos;
            while(l < r){
                int mid = (l + r + 1) >> 1;
                if(mid < pos && check(mid, pos, pos, in)) l = mid;
                else r = mid - 1;
            }
            System.out.printf("! %d\n", l);
            System.out.flush();
        } else{
            l = pos;
            r = n;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (pos < mid && check(pos, mid, pos, in)) r = mid;
                else l = mid + 1;
            }
            System.out.printf("! %d\n", l);
            System.out.flush();
        }
    }
}