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.*;

public class cf703 {

    public static Scanner sc = new Scanner(System.in);

    public static int problemA(int n, int arr[]){
        long tillNow=0;
        for(int i=0; i<n; i++){
            tillNow+=arr[i];
            if(tillNow<i) return 0;
            tillNow-=i;
        }
        return 1;
    }

    public static int binSrch(int arr[], int T){
        //get index k of arr which is last element arr[k]<T and arr[k+1]>=T
        int n = arr.length;
        int start = 0, end=n-1;

        while(start<end){
            int mid = (start+end)/2;
            if(arr[mid]<T && arr[mid+1]>=T) return mid;
            else if(arr[mid]>=T) end=mid-1;
            else start=mid;
        }
        return (end==0)?end:-1;
    }

    public static long calcD(int n, int id, long[] pr, int v){
       long sum=0;
        if(id>=0){
            sum = Math.abs((id+1)*v-pr[id])+Math.abs((pr[n-1]-pr[id]) - v* (n-1-id));
        }
        else {
            sum= Math.abs((pr[n-1]) - v* (n));
        }
        return sum;
    }

    public static int ask(int l, int r){
        System.out.println("? "+l+" "+r);
        int index = sc.nextInt();
        return index;
    }

    public static long problemB(int n, int x[], int y[]){
//        int maxX = 1000000000;
//        int maxY = 1000000000;
        Arrays.sort(x);
        Arrays.sort(y);
       long res1 = x[(n/2)]-x[(n-1)/2] +1;
       long res2 = y[(n/2)]-y[(n-1)/2] +1;
       return res1*res2;
    }

    public static void problemCD(int n){
        int smaxIndex = ask(1,n);
        int start=1; int end=n;
        if(smaxIndex==1 || ask(1,smaxIndex)!=smaxIndex){
            //between smaxindex-n
            int l=smaxIndex; int r=n;
            while((r-l)>1){
                int mid = (l+r)/2;
                if(ask(smaxIndex,mid)==smaxIndex) r=mid;
                else l=mid;
            }
            System.out.println("! "+r);
            System.out.flush();
        }
        else{
            //between 1-smaxindex
            int l=1; int r=smaxIndex;
            while((r-l)>1){
                int mid = (l+r)/2;
                if(ask(mid,smaxIndex)==smaxIndex) l=mid;
                else r=mid;
            }

            System.out.println("! "+l);
            System.out.flush();
        }

    }

    public static void main(String[] args) {
        int tcases;
        tcases = sc.nextInt();
        problemCD(tcases);

    }
}