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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;

public final class C1 {
    
    public static void main(String[] args) {
        final FastScanner fs = new FastScanner();
        final int n = fs.nextInt();
        final int res1 = query(0, n - 1, fs);
        final int res2 = query(0, res1, fs);
        int lo;
        int hi;
        if (res2 == res1) {
            lo = 0;
            hi = res1;
            while (lo < hi) {
                final int mid = lo + hi + 1 >>> 1;
                if (query(mid, res1, fs) == res1) {
                    lo = mid;
                } else {
                    hi = mid - 1;
                }
            }
        } else {
            lo = res1;
            hi = n - 1;
            while (lo < hi) {
                final int mid = lo + hi >>> 1;
                if (query(res1, mid, fs) != res1) {
                    lo = mid + 1;
                } else {
                    hi = mid;
                }
            }
        }
        System.out.println("! " + (lo + 1));
        System.out.close();
    }

    private static int query(int low, int high, FastScanner fs) {
        if (low == high) {
            return -1;
        }
        low++;
        high++;
        System.out.println("? " + low + ' ' + high);
        return fs.nextInt() - 1;
    }

    static final class Utils {
        public static void shuffleSort(int[] x) {
            shuffle(x);
            Arrays.sort(x);
        }

        public static void shuffleSort(long[] x) {
            shuffle(x);
            Arrays.sort(x);
        }

        public static void shuffle(int[] x) {
            final Random r = new Random();

            for (int i = 0; i <= x.length - 2; i++) {
                final int j = i + r.nextInt(x.length - i);
                swap(x, i, j);
            }
        }

        public static void shuffle(long[] x) {
            final Random r = new Random();

            for (int i = 0; i <= x.length - 2; i++) {
                final int j = i + r.nextInt(x.length - i);
                swap(x, i, j);
            }
        }

        public static void swap(int[] x, int i, int j) {
            final int t = x[i];
            x[i] = x[j];
            x[j] = t;
        }

        public static void swap(long[] x, int i, int j) {
            final long t = x[i];
            x[i] = x[j];
            x[j] = t;
        }

        private Utils() {}
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

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

        int[] nextIntArray(int n) {
            final int[] a = new int[n];
            for (int i = 0; i < n; i++) { a[i] = nextInt(); }
            return a;
        }

        long[] nextLongArray(int n) {
            final long[] a = new long[n];
            for (int i = 0; i < n; i++) { a[i] = nextLong(); }
            return a;
        }
    }
}