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 ll long long
#define double long double
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()

const int INF = 2e16;

void solve() {
	int n;
	cin >> n;
	
	int id;
	cout << "? 1 " << n << endl;
	cin >> id;
	
	int l = 0;
	int r = n;
	while (r - l > 1) {
		int m = (l + r) >> 1;
		int ll = max(1LL, id - m);
		int rr = min(n, id + m);
		cout << "? " << ll << " " << rr << endl;
		int x;
		cin >> x;
		if (x == id) r = m;
		else l = m;
	}
	
	int ans = 0;
	if (id - r <= 0) {
		ans = id + r;
	}
	else if (id + r > n) {
		ans = id - r;
	}
	else {
		cout << "? " << id - r << " " << id << endl;
		int x;
		cin >> x;
		if (x == id) {
			ans = id - r;
		}
		else {
			ans = id + r;
		}
	}
	cout << "! " << ans << endl;
}
	

signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	srand(time(0));
	int t = 1;
	//cin >> t;
	while (t--) solve();
}