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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;

public class problemC {
    static class Solution {
        HashMap<Point, Integer> map = new HashMap<>();

        int query(int l, int r) {
            Point p = new Point(l, r);
            if (map.containsKey(p)) return map.get(p);
            System.out.println("? " + l + " " + r);
            System.out.flush();
            int ans = fs.nextInt();
            map.put(p, ans);
            return ans;
        }

        void solve() {
            int n = fs.nextInt();
            int second = query(1, n);
            int left = second == 1 ? -1 : query(1, second);
            if (left == second) { // search left.
                int ans = -1;
                for (int low=1, high=second-1; low <= high; ) {
                    int mid = (low+high)/2;
                    int got = query(mid, second);
                    if (got == second) { // contains max.
                        ans = Math.max(ans, mid);
                        low = mid + 1;
                    } else { // does not contain max.
                        high = mid - 1;
                    }
                }
                print(ans);
                return;
            } else { // search right.
                int ans = n+1;
                for (int low=second+1, high=n; low <= high ; ) {
                    int mid = (low+high)/2;
                    int got = query(second, mid);
                    if (got == second) { // contains max.
                        high = mid - 1;
                        ans = Math.min(ans, mid);
                    } else { // does not contain max.
                        low = mid + 1;
                    }
                }
                print(ans);
                return;
            }
        }

        void print(int ans) {
            System.out.println("! " + ans);
            System.out.flush();
        }
    }

    static class Point implements Comparable<Point> {
        int x, y;

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

        @Override
        public int compareTo(Point other) {
            if (this.x == other.x) return Integer.compare(this.y, other.y);
            return Integer.compare(this.x, other.x);
        }

        @Override
        public boolean equals(Object o) {
            if (!(o instanceof Point)) return false;
            Point other = (Point)o;
            return this.x == other.x && this.y == other.y;
        }

        @Override
        public int hashCode() {
            return Arrays.hashCode(new int[]{this.x , this.y});
        }
    }

    public static void main(String[] args) throws Exception {
        Solution solution = new Solution();
        solution.solve();
    }

    static void debug(Object... O) {
        System.err.println("DEBUG: " + Arrays.deepToString(O));
    }

    private static FastScanner fs = new FastScanner();

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

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

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

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

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

        ArrayList<Integer> readArrayList(int n) {
            ArrayList<Integer> a = new ArrayList<>(n);
            for (int i = 0 ; i < n; i ++ ) a.add(fs.nextInt());
            return a;
        }

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

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

        String nextString() {
            return next();
        }
    }
}