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:
//package Current;

import java.util.*;

public class Main {
    static final long mod = (long) 1e9 + 7;

    static class pair {
        int x, y;

        public pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }

    public static void main(String args[]) {

        Scanner sc = new Scanner(System.in);
        int t = 1;
        // t = sc.nextInt();
        while (t-- > 0) {
            // Start code
            int n = sc.nextInt();

            int ans = 0;

            int smax = ask(1, n, sc);

            if (smax == 1 || ask(1, smax, sc) != smax) {
                int l = smax + 1, r = n;
                while (l <= r) {
                    int mid = l + (r - l) / 2;
                    int p = ask(smax, mid, sc);
                    if (p == smax) {
                        ans = mid;
                        r = mid - 1;
                    } else
                        l = mid + 1;
                }
            } else {
                int l = 1, r = smax - 1;
                while (l <= r) {
                    int mid = l + (r - l) / 2;
                    int p = ask(mid, smax, sc);
                    if (p == smax) {
                        ans = mid;
                        l = mid + 1;
                    } else
                        r = mid - 1;
                }
            }

            println("! " + ans);
        }
        sc.close();
    }

    static int ask(int l, int r, Scanner sc) {
        println("? " + l + " " + r);
        System.out.flush();

        int x = sc.nextInt();
        return x;
    }

    static void print(Object o) {
        System.out.print(o + " ");
    }

    static void println(Object o) {
        System.out.println(o);
    }

    static void swap(int arr[], int i, int j) {
        int temp = arr[i];
        arr[i] = arr[j];
        arr[j] = temp;
    }

    static ArrayList<Long> factorial(int num) {
        ArrayList<Long> fac = new ArrayList<>();
        fac.add((long) 0);
        fac.add((long) 1);

        for (int i = 2; i < num; i++) {
            fac.add((fac.get(i - 1) * i) % mod);
        }

        return fac;
    }

    static long ncr(long x, long y, ArrayList<Long> fac) {
        if (y >= x)
            return (long) 1;

        long res = fac.get((int) x);
        long z = (fac.get((int) y) * fac.get((int) (x - y))) % mod;
        z = modInv(z);
        res = (res * z) % mod;
        return res;
    }

    static long modInv(long x) {
        return modExpo(x, mod - 2);
    }

    static long modExpo(long x, long y) {
        long res = 1;
        x = x % mod;
        while (y > 0) {
            if (y % 2 == 1)
                res = (res * x) % mod;
            y = y / 2;
            x = (x * x) % mod;
        }
        return res;
    }

    static int lowerBound(int n, long[] arr, long value) {
        int res = (int) 1e7;
        int l = 0, r = n - 1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            if (arr[mid] >= value) {
                res = mid;
                r = mid - 1;
            } else
                l = mid + 1;
        }
        return res;
    }
}