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.util.*;

import javax.sound.midi.Track;

import java.io.*;

public class tr0 {
	static PrintWriter out;
	static StringBuilder sb;
	static long mod = (long) 1e9 + 7;
	static long inf = (long) 1e16;
	static int n, m;
	static ArrayList<Integer>[] ad;
	static int[][] remove, add;
	static int[][] memo, memo1[];
	static boolean vis[];
	static long[] f, inv, ncr[];
	static HashMap<Integer, Integer> hm;
	static int[] pre, suf, Smax[], Smin[];
	static int idmax, idmin;
	static ArrayList<Integer> av;
	static HashMap<Integer, Integer> mm;
	static boolean[] msks;
	static int[] lazy[], lazyCount;
	static int[] a, dist;
	static char[][] g;
	static int ave;
	static ArrayList<Integer> gl;
	static int[][] arr;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int n = sc.nextInt();
		int pos = query(1, n);
		if (n == 2) {
			System.out.println("! " + (n - pos + 1));
			System.out.flush();
			return;
		}
		int ans = 0;
		if (pos != n) {
			int l = query(pos, n);
			if (l == pos) {
				int sign = 1;
				int dest = pos;
				for (int p = 17; p >= 0; p--) {
					if ((dest + sign * (1 << p)) <= n) {
						dest += sign * (1 << p);
						int q = query(pos, dest);
						if (q == pos) {
							ans = dest;
							sign = -1;
						} else {
							sign = 1;
						}
					}
				}
			} else {
				int sign = -1;
				ans = pos;
				int dest = pos;
				for (int p = 17; p >= 0; p--) {
					if ((dest + sign * (1 << p)) >= 1) {
						dest += sign * (1 << p);
						int q = query(dest, pos);
						if (q == pos) {
							ans = dest;
							sign = 1;
						} else {
							sign = -1;
						}
					}
				}
			}
		} else {
			int sign = -1;
			int dest = pos;
			for (int p = 17; p >= 0; p--) {
				if ((dest + sign * (1 << p)) >= 1) {
					dest += sign * (1 << p);
					int q = query(dest, pos);
					if (q == pos) {
						ans = dest;
						sign = 1;
					} else {
						sign = -1;
					}
				}
			}
		}
		System.out.println("! " + ans);
		System.out.flush();
		// out.flush(); 1 2 3
	}

	static int query(int l, int r) throws IOException {
		Scanner sc1 = new Scanner(System.in);
		System.out.println("? " + l + " " + r);
		System.out.flush();
		return sc1.nextInt();
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

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

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

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

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

		public int[] nextArrIntSorted(int n) throws IOException {
			int[] a = new int[n];
			Integer[] a1 = new Integer[n];
			for (int i = 0; i < n; i++)
				a1[i] = nextInt();
			Arrays.sort(a1);
			for (int i = 0; i < n; i++)
				a[i] = a1[1].intValue();
			return a;
		}

		public long[] nextArrLongSorted(int n) throws IOException {
			long[] a = new long[n];
			Long[] a1 = new Long[n];
			for (int i = 0; i < n; i++)
				a1[i] = nextLong();
			Arrays.sort(a1);
			for (int i = 0; i < n; i++)
				a[i] = a1[1].longValue();
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}