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.*;
import java.math.BigInteger;
import java.util.*;
import java.util.concurrent.TimeUnit;

public class c1486 implements Runnable{
	
    public static void main(String[] args) {
    	try{
            new Thread(null, new c1486(), "process", 1<<26).start();
        }
        catch(Exception e){
            System.out.println(e);
        }
    }
	public void run() {
		FastReader scan = new FastReader();
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out), true);
		//PrintWriter out = new PrintWriter("file.out");
		Task solver = new Task();
		//int t = scan.nextInt();
		int t = 1;
		for(int i = 1; i <= t; i++) solver.solve(i, scan, out);
		out.close();
	}

	static class Task {
		static final int oo = Integer.MAX_VALUE;
		static final long OO = Long.MAX_VALUE;

		public void solve(int testNumber, FastReader sc, PrintWriter out) {
			int N = sc.nextInt();
			
			out.println("? 1 " + N);
			int secondMax = sc.nextInt();
			
			boolean left = false;
			if(secondMax == 1) {
				left = false;
			} else if (secondMax == N) {
				left = true;
			} else {
				out.println("? 1 " + secondMax);
				int t = sc.nextInt();
				if(t == secondMax) {
					left = true;
				}
			}
			
			
			int L = 0, R = (N - secondMax);
			if(left) {
				R = secondMax - 1;
			}
			while(L < R - 1) {
				int M = (L + R) / 2;
				
				int qM = secondMax + M;
				if(left) {
					qM = secondMax - M;
				}
				
				out.println("? " + Math.min(qM, secondMax) + " " + Math.max(qM, secondMax));
				int t = sc.nextInt();
				
				if(t == secondMax) {
					// within the range queried
					R = M;
				} else {
					L = M+1;
				}
			}
			
			if(L != R) {
				if(left) {
					int tL = secondMax - R;
					int tR = secondMax - L;
					L = tL; R = tR;
				} else {
					L = secondMax + L;
					R = secondMax + R;
				}
				out.println("? " + L + " " + R);
				if(sc.nextInt() == L) {
					out.println("! "+ R);
				} else {
					out.println("! " + L);
				}
				return;
			}
			
			if(left)
				out.println("! " + (secondMax - L));
			else
				out.println("! " + (secondMax + L));
		}
		
	}
	static long modInverse(long N, long MOD) {
		return binpow(N, MOD - 2, MOD);
	}
	static long modDivide(long a, long b, long MOD) {
		a %= MOD;
		return (binpow(b, MOD-2, MOD) * a) % MOD;
	}
	static long binpow(long a, long b, long m) {
		a %= m;
		long res = 1;
		while (b > 0) {
			if ((b & 1) == 1)
				res = res * a % m;
			a = a * a % m;
			b >>= 1;
		}
		return res;
	}
	static int[] reverse(int a[]) 
    { 
        int[] b = new int[a.length]; 
        for (int i = 0, j = a.length; i < a.length; i++, j--) { 
            b[j - 1] = a[i]; 
        } 
        
        return b;
    }
	static long[] reverse(long a[]) 
    { 
        long[] b = new long[a.length]; 
        for (int i = 0, j = a.length; i < a.length; i++, j--) { 
            b[j - 1] = a[i]; 
        } 
        
        return b;
    }
	
	static void shuffle(Object[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			Object temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}
	
	static void shuffle(int[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			int temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}
	static void shuffle(long[] a) {
		Random get = new Random();
		for (int i = 0; i < a.length; i++) {
			int r = get.nextInt(a.length);
			long temp = a[i];
			a[i] = a[r];
			a[r] = temp;
		}
	}
	
	static class tup implements Comparable<tup>, Comparator<tup>{
		int a, b;
		tup(int a,int b){
			this.a=a;
			this.b=b;
		}
		public tup() {
		}
		@Override
		public int compareTo(tup o){
			return Integer.compare(b,o.b);
		}
		@Override
		public int compare(tup o1, tup o2) {
			return Integer.compare(o1.b, o2.b);
		}
		
		@Override
	    public int hashCode() {
			return Objects.hash(a, b);
	    }
 
	    @Override
	    public boolean equals(Object obj) {
	    	if (this == obj)
                return true;
	    	if (obj == null)
                return false;
	    	if (getClass() != obj.getClass())
                return false;
	    	tup other = (tup) obj;
	    	return a==other.a && b==other.b;
	    }
	    
	    @Override
	    public String toString() {
	    	return a + " " + b;
	    }
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		public FastReader(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(new File(s)));
		}

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

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

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

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

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
		
		int[] readArray(int size) {
			int[] a = new int[size];
			for(int i = 0; i < size; i++) {
				a[i] = nextInt();
			}
			return a;
		}
		
		long[] readLongArray(int size) {
			long[] a = new long[size];
			for(int i = 0; i < size; i++) {
				a[i] = nextLong();
			}
			return a;
		}
	}
	
	static void dbg(int[] arr) {
		System.out.println(Arrays.toString(arr));
	}
	static void dbg(long[] arr) {
		System.out.println(Arrays.toString(arr));
	}
	static void dbg(boolean[] arr) {
		System.out.println(Arrays.toString(arr));
	}

	static void dbg(Object... args) {
        for (Object arg : args)
            System.out.print(arg + " ");
        System.out.println();
    }
}