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 endl '\n'
#define fi first
#define se second
#define forn(i,n) for (int i = 0; i < n; i++)
#define forr(i,a,b) for (int i = a; i <= b; i++)
#define all(v) v.begin(), v.end()
#define pb push_back
#define fastIO(); ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define SZ(s) int(s.size())
#define var(x) #x << " = " << x

using namespace std;

typedef long long lli;
typedef long double ld;
typedef pair<lli, lli> plli;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<vector<lli>> vvlli;
typedef vector<lli> vlli;
typedef vector<pii> vpii;

map<pii, int> dp;

int query(int l, int r){
	if(l == r) return -1;
	if(dp.count({l, r})) return dp[{l, r}];
	cout << "? " << l << ' ' << r << endl;
	cout.flush();
	int ans;
	cin >> ans;
	return dp[{l, r}] = ans;
}

int main () {
	fastIO();
	int n;
	cin >> n;
	int last_q = query(1, n);
	int l, r;
	l = 1; r = n;
	while(l < r){
		int m = (l + r) >> 1;
		if(last_q <= m){
			int q = query(1, m);
			if(q == last_q)
				r = m;
			else
				l = m + 1;
		}
		else{
			int q = query(m + 1, n);
			if(q == last_q)
				l = m + 1;
			else
				r = m;
		}
	}
	cout << "! " << r << endl;
	return 0;
}