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;
// #include <atcoder/all>
// using namespace atcoder;

#define rep(i,n) for (int i = 0; i < (n); ++i)
#define STRING(str) #str
#define ll long long
#define ld long double

template <typename T>
void print(T x, string name) {
	cout << name+": " << x << endl;
}

template <typename T >
void print_vec_1d(vector<T> x, string name) {
	cout << name << endl;
	cout << "[ ";
	for (int i=0;i<x.size();i++) {
		cout << x[i];
		if (i != x.size()-1) cout << ", ";
		else cout << " ]" << endl;
	}
}

template <typename T >
void print_vec_2d(vector<vector<T>> x, string name) {
	cout << name << endl;
	cout << "[ ";
	for (int i=0;i<x.size();i++) {
		cout << "[ ";
		for (int j=0;j<x[i].size();j++) {
			cout << x[i][j];
			if (j != x[i].size()-1) cout << ", ";
			else {
				cout << " ]";
				if (!((i == x.size()-1) && (j == x[i].size()-1))) {
					cout << "," << endl;
				}
			}
		}
	}
	cout << " ]" << endl;
}

template <typename T >
void print_vec_3d(vector<vector<vector<T>>> x, string name) {
	cout << name << endl;
	cout << "[ ";
	for (int i=0;i<x.size();i++) {
		cout << "[ ";
		for (int j=0;j<x[i].size();j++) {
			cout << "[ ";
			for (int k=0;k<x[i][j].size();k++) {
				cout << x[i][j][k];
				if (k != x[i][j].size()-1) cout << ", ";
				else {
					cout << " ]";
					if (!((i == x.size()-1) && (j == x[i].size()-1) && (k == x[i][j].size()-1))) {
						cout << "," << endl;
					}
				}
			}
			if (j != x[i].size()-1) cout << ", ";
			else {
				cout << " ]";
				if (!((i == x.size()-1) && (j == x[i].size()-1))) {
					cout << "," << endl;
				}
			}
		}
	}
	cout << " ]" << endl;
}

int main() {
	// ios::sync_with_stdio(0);
	// cin.tie(0);

	int n;
	cin >> n;
	int ind_2;
	cout << "? 1 " << n << "\n";
	cin >> ind_2;
	ind_2--;

	int a1, a2;
	if (ind_2 != 0 && ind_2 != n-1) {
		cout << "? 1 " << ind_2+1 << "\n";
		cin >> a1;
		a1--;

		int a2;
		cout << "? " << ind_2+1 << " " << n << "\n";
		cin >> a2;
		a2--;
	}

	bool left_flag;
	if (ind_2 == n-1) {
		left_flag = true;
	} else if (ind_2 == 0) {
		left_flag = false;
	} else if (ind_2 == a1) {
		left_flag = true;
	} else { // ind_2 == a2
		left_flag = false;
	}
	if (left_flag) {
		// ? ? ? 1 ? ? ? 2 ? ? ? ?
		int ok = 0;
		int ng = ind_2;
		while (abs(ok-ng) > 1) {
			int mid = (ok+ng)/2;
			int b;
			cout << "? " << mid+1 << " " << ind_2+1 << "\n";
			cin >> b;
			b--;
			if (b == ind_2) {
				ok = mid;
			} else {
				ng = mid;
			}
		}
		cout << "! " << ok+1 << "\n";
	} else {
		// ? ? ? ? ? ? ? 2 ? 1 ? ?
		int ok = n-1;
		int ng = ind_2;
		while (abs(ok-ng) > 1) {
			int mid = (ok+ng)/2;
			int b;
			cout << "? " << ind_2+1 << " " << mid+1 << "\n";
			cin >> b;
			b--;
			if (b == ind_2) {
				ok = mid;
			} else {
				ng = mid;
			}
		}
		cout << "! " << ok+1 << "\n";
	}
	return 0;
}