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 javax.print.DocFlavor;
import javax.swing.text.html.parser.Entity;
import java.awt.*;
import java.io.*;
import java.lang.reflect.Array;
import java.util.*;
import java.util.List;
import java.util.stream.Collectors;

public class Main {

    static FastScanner sc;
    static PrintWriter pw;
    static final long INF = 200000000000000l;
    static final int MOD = 1000000007;
    static final int N = 1005;

    static int mul(int a, int b) {
        return (int) ((a * 1l * b) % MOD);
    }


    public static int pow(int a, int n) {
        int res = -1;
        while(n != 0) {
            if((n & 1) == 1) {
                res = mul(res, a);
            }

            a = mul(a, a);
            n >>= 1;
        }

        return res;
    }

    public static int inv(int a) {
        return pow(a, MOD-2);
    }

    public static List<Integer> getPrimes() {
        List<Integer> ans = new ArrayList<>();
        boolean[] prime = new boolean[N];
        Arrays.fill(prime, true);
        prime[0] = prime[1] = false;
        for(int i = 2; i < N; ++i) {
            if(prime[i]) {
                ans.add(i);
                if (i * 1l * i <= N)
                    for (int j = i * i; j < N; j += i)
                        prime[j] = false;
            }
        }

        return ans;
    }

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

    public static class A {
        long x;
        long y;
        A(long a, long b) {
            this.x = a;
            this.y = b;
        }
    }


    public static void solve() throws IOException {
        int n = sc.nextInt();
        pw.println("? 1 " + n);
        pw.flush();
        int second = sc.nextInt();
        int l = 0, r = 0;
        boolean left = true;
        if(second == n) {
            left = false;
            r = second;
            l = 0;
        } else if(second == 1) {
            l = second;
            r = n + 1;
        } else {
            pw.println("? " + second + " " + n);
            pw.flush();
            if(second == sc.nextInt() && second != n) {
                l = second;
                r = n + 1;
            }
            else {
                r = second;
                l = 0;
                left = false;
            }
        }
        while(r - l >= 2) {
            int mid = (l + r) / 2;
            if(left) {
                pw.println("? " + second + " " + mid);
                pw.flush();
                int x = sc.nextInt();
                if(x == second)
                    r = mid;
                else l = mid;
            } else {
                pw.println("? " + mid + " " + second);
                pw.flush();
                int x = sc.nextInt();
                if(x == second) l = mid;
                else r = mid;
            }
        }

        pw.println("! " + (left ? (l + 1) : (l)) );
    }


    public static void main(String[] args) throws IOException {
        sc = new FastScanner(System.in);
        pw = new PrintWriter(System.out);
        int xx = 1;
        while(xx > 0) {
            solve();
            xx--;
        }
        pw.close();

    }


}

class FastScanner {
    static StringTokenizer st = new StringTokenizer("");
    static BufferedReader br;

    public FastScanner(InputStream inputStream) {
        br = new BufferedReader(new InputStreamReader(inputStream));
    }

    public String next() throws IOException {
        while (!st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

}