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.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Formatter;
import java.util.Random;

/**
 * Coder: SumitRaut
 * Date: 27-02-2021 14:23
 */

public class Main {
    public static void main(String[] args) throws IOException {
        Fs = new FastReader();
        out = new PrintWriter(System.out);
        int n = Fs.nextInt();
        out.println("? " + 1 + " " + n);
        out.flush();
        int id = Fs.nextInt();
        boolean done=false;
        int l=1,r=n;
        if(id!=1) {
            out.println("? "+1+" "+id);
            out.flush();
            if(Fs.nextInt()==id) {
                done=true;
                r=id-1;
                while(l<r) {
                    int md=(l+r+1)>>1;
                    out.println("? "+md+" "+id);
                    out.flush();
                    int curif=Fs.nextInt();
                    if(curif==id) l=md;
                    else r=md-1;
                }
            }
        }
        if(!done) {
            l=id+1;
            while(l<r) {
                int md=(l+r)>>1;
                out.println("? "+id+" "+md);
                out.flush();
                int curif=Fs.nextInt();
                if(curif==id) r=md;
                else l=md+1;
            }
        }
        out.print("! " + r);
        out.flush();
        out.close();
    }

    static PrintWriter out;
    static FastReader Fs;
    static final Random random = new Random();

    static void ruffleSort(int[] a) {
        int n = a.length;
        for (int i = 0; i < n; ++i) {
            int oi = random.nextInt(n), tmp = a[oi];
            a[oi] = a[i];
            a[i] = tmp;
        }
        Arrays.sort(a);
    }

    static class FastReader {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public FastReader() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public FastReader(String file_name) throws IOException {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String nextLine() throws IOException {
            byte[] buf = new byte[64]; // 64 line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == '\n') {
                    if (cnt != 0) {
                        break;
                    } else {
                        continue;
                    }
                }
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public String next() throws IOException {
            byte[] buf = new byte[64]; // 64 line length
            int cnt = 0, c;
            while ((c = read()) != -1) {
                if (c == ' ' || c == '\n') {
                    if (cnt != 0) {
                        break;
                    } else {
                        continue;
                    }
                }
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public long nextLong() throws IOException {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (c == '.') {
                while ((c = read()) >= '0' && c <= '9') {
                    ret += (c - '0') / (div *= 10);
                }
            }
            if (neg)
                return -ret;
            return ret;
        }

        public int[] readArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; ++i)
                a[i] = nextInt();
            return a;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException {
            if (din == null)
                return;
            din.close();
        }
    }
}