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  : dante1
 * created : 18/02/2021 23:12
 *******************************************************************************/
import java.io.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
// import java.math.BigInteger;
import java.util.*;

public class c1 {
    public static void main(String[] args) {
        int n = in.nextInt(), t = query(1, n), max_idx = 0;
        if (t == query(1, t)) {
            int lo = 1, hi = t-1;
            while (lo <= hi) {
                int mid = (lo + hi) >> 1;
                if (t == query(mid, t)) {
                    max_idx = mid;
                    lo = mid+1;
                } else {
                    hi = mid-1;
                }
            }
        } else {
            int lo = t+1, hi = n;
            while (lo <= hi) {
                int mid = (lo + hi) >> 1;
                if (t == query(t, mid)) {
                    max_idx = mid;
                    hi = mid-1;
                } else {
                    lo = mid+1;
                }
            }
        }
        out.println("! " + max_idx);
        out.flush();
    }
    
    static int query(int l, int r) {
        if (l == r) {
            return 0;
        }
        out.println("? " + l + " " + r);
        out.flush();
        return in.nextInt();
    } 

    // Handle I/O
    static int MOD = (int) (1e9 + 7);
    static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        public FastScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        int[] readIntArray(int size) {
            int[] arr = new int[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextInt();
            }
            return arr;
        }

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

        long[] readLongArray(int size) {
            long[] arr = new long[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextLong();
            }
            return arr;
        }

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

        double[] readDoubleArray(int size) {
            double[] arr = new double[size];
            for (int i = 0; i < size; i++) {
                arr[i] = nextDouble();
            }
            return arr;
        }
    }
}