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.util.*;
import java.lang.*;

public class C {
    public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static long MOD = (long) (1e9 + 7);
    //static int MOD = 998244353;
    static long MOD2 = MOD * MOD;
    static FastReader sc = new FastReader();
    static int pInf = Integer.MAX_VALUE;
    static int nInf = Integer.MIN_VALUE;
    public static void main(String[] args) {
        int test = 1;
        //test = sc.nextInt();
        for (int i = 1; i <= test; i++){
            //out.print("Case #"+i+": ");
            solve();
        }
        out.flush();
        out.close();
    }
    static long oo = (long) (1e16);
    static void solve(){
        int n = sc.nextInt();
        int l = 1;
        int r = n;
        int ele = query(1,n);
        int lele = -1;
        int rele = -1;
        if(ele>l){
            lele = query(l,ele);
        }
        if(r>ele){
            rele = query(ele,r);
        }
        if(lele==ele){
            r = ele;
            l--;
            while(r-l>1){
                int m = l+(r-l)/2;
                int idx = query(m,ele);
                if(idx==ele){
                    l = m;
                }else{
                    r = m;
                }
            }
            out.println("! "+l);
            out.flush();
        }else if(rele==ele){
            l = ele;
            r++;
            while(r-l>1){
                int m = l+(r-l)/2;
                int idx = query(ele,m);
                if(idx==ele){
                    r = m;
                }else{
                    l = m;
                }
            }
            out.println("! "+r);
            out.flush();
        }


    }
    static int query(int l,int r){
        out.println("? "+l+" "+r);
        out.flush();
        return sc.nextInt();
    }
    static long nC2(long n) {
        return add((n * (n + 1)) / 2, 0);
    }
    public static long mul(long a, long b) {
        return ((a % MOD) * (b % MOD)) % MOD;
    }
    public static long add(long a, long b) {
        return ((a % MOD) + (b % MOD)) % MOD;
    }
    public static long c2(long n) {
        if ((n & 1) == 0) {
            return mul(n / 2, n - 1);
        } else {
            return mul(n, (n - 1) / 2);
        }
    }
    //Pair Class
    static class Pair implements Comparable<Pair> {
        int x;
        int y;
        public Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }
        @Override
        public int compareTo(Pair o) {
            if ((this.x == o.x)) {
                return (this.y - o.y);
            }
            return (this.x - o.x);
        }
    }
    //Shuffle Sort
    static final Random random = new Random();
    static void ruffleSort(int[] a) {
        int n = a.length;//shuffle, then sort
        for (int i = 0; i < n; i++) {
            int oi = random.nextInt(n), temp = a[oi];
            a[oi] = a[i];
            a[i] = temp;
        }
        Arrays.sort(a);
    }
    //Brian Kernighans Algorithm
    static long countSetBits(long n) {
        if (n == 0) return 0;
        return 1 + countSetBits(n & (n - 1));
    }
    //Euclidean Algorithm
    static long gcd(long A, long B) {
        if (B == 0) return A;
        return gcd(B, A % B);
    }
    //Modular Exponentiation
    static long fastExpo(long x, long n) {
        if (n == 0) return 1;
        if ((n & 1) == 0) return fastExpo((x * x) % MOD, n / 2) % MOD;
        return ((x % MOD) * fastExpo((x * x) % MOD, (n - 1) / 2)) % MOD;
    }
    //AKS Algorithm
    static boolean isPrime(long n) {
        if (n <= 1) return false;
        if (n <= 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        for (int i = 5; i <= Math.sqrt(n); i += 6)
            if (n % i == 0 || n % (i + 2) == 0) return false;
        return true;
    }
    public static long modinv(long x) {
        return modpow(x, MOD - 2);
    }
    public static long modpow(long a, long b) {
        if (b == 0) {
            return 1;
        }
        long x = modpow(a, b / 2);
        x = (x * x) % MOD;
        if (b % 2 == 1) {
            return (x * a) % MOD;
        }
        return x;
    }

    public static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        //it reads the data about the specified point and divide the data about it ,it is quite fast
        //than using direct

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

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception r) {
                    r.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());//converts string to integer
        }

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

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (Exception r) {
                r.printStackTrace();
            }
            return str;
        }
    }
}