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:
#define LOCAL09
#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define debug(msg) cout << (msg) << endl
#define FOR(i, a, b) for (int i=a; i<=(b); i++)
#define ROF(i, a, b) for (int i=a; i>=(b); i--)
#define pb(x) push_back(x) 
#define ALL(x) x.begin(), x.end()
#define RALL(x) x.rbegin(), x.rend()
#define sz(x) (ll)x.size()
#define JUDGE(x) if(x) cout << "YES\n"; else cout << "NO\n";
using namespace std;
typedef long long ll;
const ll mod = 1e8+7;
const int maxn = 20;
ll x[2000], y[2000];
int main()
{
	#ifdef LOCAL
		freopen("in.txt", "r", stdin);
    	freopen("out.txt", "w", stdout);
	#endif
	int T = 1;
	// cin >> T;
	while(T--){
		int n;
		cin >> n;
		int pos, t=-1;
		cout << "? 1 " << n << endl;
		cin >> pos;
		if(pos!=1){
			cout << "? 1 " << pos << endl;
		cin >> t;
		}
		
		if(t!=-1 && t==pos){
			int l = 1, r = pos;
			while(r-l>1){
				int mid = (l+r)>>1;
				cout << "? " << mid << " " << pos << endl;
				cin >> t;
				if(t==pos) l = mid;
				else r = mid;
			}
			cout << "! " << l << endl;
		}
		else{
			int l = pos, r = n;
			while(r-l>1){
				int mid = (l+r)>>1;
				cout << "? " << pos << " " << mid << endl;
				cin >> t;
				if(t==pos) r = mid;
				else l = mid;
			}
			cout << "! " << r << endl;
		}
	}
    return 0;
}