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>
#define IO                        \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define mem(n, x) memset(n, x, sizeof(n))
#define lp(i, n) for (int i = 0; i < n; i++)
#define lp1(i, n) for (int i = 1; i <= n; i++)
#define ull unsigned long long
#define ll long long
#define re return
#define endl '\n'
using namespace std;
const double PI = acos(-1);
const int N = 5e5 + 5, M = 5e5 + 5, mod = 1e9 + 7, OO = 0x3f3f3f3f;
ll tc , n,m;
ll arr[N];
ll a[N];
int ask(int  a, int b)
{
	if (a >= b)
		re - 1;
	cout << "? " << a << "  " << b << endl;
	cout.flush();
	int x; 
	cin >> x;
	re x;
}
int main()
{
	IO;
	cin >> n; 
	ll ans = 0; 
	int  mx = ask(1, n);
	if (ask(1, mx) == mx && mx != 1)
	{
		ll st = 1, mid, en =mx-1;
		while (st <= en)
		{
			mid = (st + en) / 2;
			if (ask(mid, mx) == mx)
			{
				st = mid + 1;
				ans = mid;
			}
			else
				en = mid - 1;
		}
	}
	else
	{
		ll st = mx+1 , mid, en = n;
		while (st <= en)
		{
			mid = (st + en) / 2;
			if (ask(mx, mid) == mx)
			{
				en = mid - 1;
				ans = mid;
			}
			else
			{
				st = mid + 1;
			}
	    }
	}
	cout << "! " << ans << endl;
	
	
}

/*


*/