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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;

#define int long long
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

int query(int l, int r) {
	if (l == r) {
		return -1;
	}

	cout << "? " << l << " " << r << endl;
	cout.flush();
	int ans;
	cin >> ans;
	return ans;
}

void solve() {
	int n;
	cin >> n;

	int idx =  query(1, n);
	int tmpIdx = query(1, idx);

	int ans;
	int l = 1, r = n;
	if (tmpIdx != idx) {
		l = idx + 1, r = n;
		while (l <= r) {
			int mid = (l + r) / 2;

			if (query(idx, mid) != idx) {
				l = mid + 1;
			} else {
				ans = mid;
				r = mid - 1;
			}
		}
	} else {
		l = 1, r = idx - 1;
		while (l <= r) {
			int mid = (l + r) / 2;

			if (query(mid, idx) != idx) {
				r = mid - 1;
			} else {
				ans = mid;
				l = mid + 1;
			}
		}
	}

	cout << "! " << ans << endl;
	cout.flush();
	// answer(ans);

	return;
}

signed main() {
	FIO;

	int tc(1);
	// cin >> tc;

	for (int i = 0; i < tc; i++) {
		// cout << "tc #" << i + 1 << " : ";
		solve();
	}

	return 0;
}