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.*;
import java.util.*;

public class C {
    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;
        }
    }

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

    private static int[] rai(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextInt();
        }
        return arr;
    }

    private static int[][] rai(int n, int m) {
        int[][] arr = new int[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextInt();
            }
        }
        return arr;
    }

    private static long[] ral(int n) {
        long[] arr = new long[n];
        for (int i = 0; i < n; i++) {
            arr[i] = s.nextLong();
        }
        return arr;
    }

    private static long[][] ral(int n, int m) {
        long[][] arr = new long[n][m];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                arr[i][j] = s.nextLong();
            }
        }
        return arr;
    }

    private static int ri() {
        return s.nextInt();
    }

    private static long rl() {
        return s.nextLong();
    }

    private static String rs() {
        return s.next();
    }

    static long gcd(long a,long b)
    {
        if(b==0)
        {
            return a;
        }
        return gcd(b,a%b);
    }

    static boolean isPrime(int n) {
        //check if n is a multiple of 2
        if (n % 2 == 0) return false;
        //if not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (n % i == 0)
                return false;
        }
        return true;
    }
    static int MOD=1000000007;

    static long mpow(long base,long pow)
    {
        long res=1;
        while(pow>0)
        {
            if(pow%2==1)
            {
                res=(res*base)%MOD;
            }
            pow>>=1;
            base=(base*base)%MOD;
        }
        return res;
    }

    static public int minimumSize(int[] nums, int m) {
        PriorityQueue<Integer> pq=new PriorityQueue<>(Collections.reverseOrder());
        for(int i:nums)
        {
            pq.add(i);
        }
        System.out.println(pq);
        for(int i=0;i<m;i++)
        {
            int val=pq.remove();

            if(val==1)
            {
                return 1;
            }
            int f=val/2;
            int s=val-f;
            pq.add(f);
            pq.add(s);
            System.out.println(pq);
        }
        return pq.remove();
    }

    public static void main(String[] args) {
//        StringBuilder ans = new StringBuilder();
//        int t = ri();
        int t = 1;
        while (t-- > 0)
        {
            int n=ri();
            int l=1,r=n;
            boolean f=false;
            int count=0;
            out.println("? 1 "+n);
            out.flush();
            count++;
            int secMax=ri();
            boolean before=false;

            if(secMax==1)
            {
                l=2;
            }
            else if(secMax==n)
            {
                r=n-1;
                before=true;
            }
            else
            {
                out.println("? 1 "+secMax);
                out.flush();
                count++;
                int ind=ri();
                if(ind==secMax)
                {
                    r=secMax-1;
                    before=true;
                }
                else {
                    l=secMax+1;
                }
            }

            int res=-1;
            while(l<=r)
            {
                int mid=(l+r)/2;
                if(before) {
                    out.println("? " + mid + " " +secMax);
                    out.flush();
                    count++;
                    int ind=ri();
                    if(ind==secMax)
                    {
                        res=mid;
                        l=mid+1;
                    }
                    else {
                        r=mid-1;
                    }
                }
                else
                {
                    out.println("? " + secMax + " " +mid);
                    out.flush();
                    count++;
                    int ind=ri();
                    if(ind==secMax)
                    {
                        res=mid;
                        r=mid-1;
                    }
                    else {
                        l=mid+1;
                    }
                }
            }

            out.println("! " + res);
            out.flush();

//            System.out.println(count);


        }
//        out.print(ans.toString());
//        out.flush();
    }
}