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:


//@author->.....future_me......//

//..............Learning.........//

/*Compete against yourself*/

import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.BigInteger;


public class C {

  // >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Code Starts <<<<<<<<<<<<<<<<<<<<<<<<<<<<< //

  static FastReader  sc  = new FastReader();
  static PrintWriter out = new PrintWriter(System.out);

  public static void main(String[] args) throws Exception {
    try {
//      int t = sc.nextInt();
//      while (t-- > 0)
        C.go();
      out.flush();

    } catch (Exception e) {
      return;
    }
  }

  static void go() {
  	int n=sc.nextInt();
  	int x=ask(1,n);
  	int ans=0;
  	if(x==ask(1,x)) {
  		int l=1;int r=x-1;
  		while(l<=r) {
  			int mid=(l+r)/2;
  			if(ask(mid,x)==x) {
  				ans=mid;
  				l=mid+1;
  			}else {
  		r=mid-1;
  			}
  		}
  	}else {
  		int l=x+1;int r=n;
  		while(l<=r) {
  			int mid=(l+r)/2;
  			if(ask(x,mid)==x) {
  				ans=mid;
  				r=mid-1;
  			}else {
  				l=mid+1;
  			}
  		}
  	}
  	out.println("! "+ans);
  	out.flush();


  	
 }
  static int ask(int l ,int r) {
  	if(l==r) {
  		return -1;
  	}
  	out.println("? "+l+" "+r);
  	out.flush();
  	int x=sc.nextInt();
  	return x;
  }

  // >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Code Ends <<<<<<<<<<<<<<<<<<<<<<<<<<<<< //


  static int mod = (int) 1e9 + 7;

  // Returns nCr % p using Fermat's
  // little theorem.
  static long fac[];

  static long ncr(int n, int r, int p) {

    if (n < r)
      return 0;
    // Base case
    if (r == 0)
      return 1;

    // Fill factorial array so that we
    // can find all factorial of r, n
    // and n-r

    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) % p;
  }

  private static long modInverse(long l, int p) {
    
    return pow(l,p-2);
  }

  static void sort(long[] a) {
    ArrayList<Long> aa = new ArrayList<>();
    for (long i : a) {
      aa.add(i);
    }
    Collections.sort(aa);
    for (int i = 0; i < a.length; i++)
      a[i] = aa.get(i);
  }

  static long pow(long x, long y) {
    long res = 1l;
    while (y != 0) {
      if (y % 2 == 1) {
        res = x * res % mod;
      }
      y /= 2;
      x = (x * x) % mod;
    }
    return res;
  }
//>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Fast IO <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< //

  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;
    }
  }

}
//use this for double quotes inside string
// " \"asdf\""