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 static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.io.*;
import java.util.*;

public class B
{
    static int n;
    static int[] arr;
    static char[] s;
    public static void main(String[] args) throws IOException
    {
        f = new Flash();
        out = new PrintWriter(System.out);
        
        int T = 1; //ni();
        for(int tc = 1; tc <= T; tc++){
            n = ni(); fn();
        }
        
        out.flush(); out.close();
    }
    
    static void fn()
    {
    	//we cannot query [L, L] or [R, R]
    	int pos = query(1, n);
    	int left = 1; 
    	if(pos > 1) left = query(1, pos);
    	int right = n; 
    	if(pos < n) right = query(pos, n);
    	if(left == pos && pos > 1) {
    		int l = 1, r = pos-1;
    		while(l < r) {
    			int mid = l + (r - l + 1) / 2;
    			if(query(mid, pos) == pos) l = mid;
    			else r = mid - 1;
    		}
    		
    		sop("! " + l);
    	}
    	
    	else {
    		int l = pos+1, r = n;
    		while(l < r) {
    			int mid = l + (r - l) / 2;
    			if(query(pos, mid) == pos) r = mid;
    			else l = mid + 1;
    		}
    		
    		sop("! " + l);
    	}
    }
    
    static int query(int l, int r) {
    	sop("? " + l + " " + r);
    	out.flush();
    	return ni();
    }
    
    static Flash f;
    static PrintWriter out;
    static final long mod = (long)1e9+7;
    static final long inf = Long.MAX_VALUE;
    static final int _inf = Integer.MAX_VALUE;
    static final int maxN = (int)5e5+5;
    static long[] fact, inv;
    
    static void sort(int[] a){
        List<Integer> A = new ArrayList<>();
        for(int i : a) A.add(i);
        Collections.sort(A);
        for(int i = 0; i < A.size(); i++) a[i] = A.get(i);
    }
    
    static void sort(long[] a){
        List<Long> A = new ArrayList<>();
        for(long i : a) A.add(i);
        Collections.sort(A);
        for(int i = 0; i < A.size(); i++) a[i] = A.get(i);
    }
    
    static void print(int[] a){
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < a.length; i++) sb.append(a[i] + " ");
        sop(sb);
    }
    
    static void print(long[] a){
        StringBuilder sb = new StringBuilder();
        for(int i = 0; i < a.length; i++) sb.append(a[i] + " ");
        sop(sb);
    }
    
    static int swap(int itself, int dummy){return itself;}
    static long swap(long itself, long dummy){return itself;}
    static void sop(Object o){out.println(o);}
    static int ni(){return f.ni();}
    static long nl(){return f.nl();}
    static double nd(){return f.nd();}
    static String next(){return f.next();}
    static String ns(){return f.ns();}
    static char[] nc(){return f.nc();}
    static int[] arr(int len){return f.arr(len);}
    static int gcd(int a, int b){if(b == 0) return a; return gcd(b, a%b);}
    static long gcd(long a, long b){if(b == 0) return a; return gcd(b, a%b);}
    static int lcm(int a, int b){return (a*b)/gcd(a, b);}
    static long lcm(long a, long b){return (a*b)/gcd(a, b);}
    
    static class Flash
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        
        String next(){
            while(!st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch(IOException e){
                    e.printStackTrace();
                }
            }
            
            return st.nextToken();
        }
        
        String ns(){
            String s = new String();
            try{
                s = br.readLine().trim();
            }catch(IOException e){
                e.printStackTrace();
            }
            
            return s;
        }
        
        int[] arr(int n){
            int[] a = new int[n];
            for(int i = 0; i < n; i++) a[i] = ni();
            return a;
        }
        
        char[] nc(){return ns().toCharArray();}
        int ni(){return Integer.parseInt(next());}
        long nl(){return Long.parseLong(next());}
        double nd(){return Double.parseDouble(next());}
    }
}