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 <cmath>
#include <stdio.h>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int INTINF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;

inline void file () {
	freopen("test.in", "r", stdin);
	freopen("test.out", "w", stdout);
}

int read () {
	int ans = 0, f = 1; char ch = getchar();
	while(!isdigit(ch)) { if(!(ch ^ '-')) f = -1; ch = getchar(); }
	while(isdigit(ch)) ans = ans * 10 + ch - '0', ch = getchar();
	return ans * f;
}
void prt (int x) {
	if(x > 9) prt(x / 10);
	putchar(48 + x % 10);
}

void print (int x, char ch = 0) {
	if(x < 0) x = -x, putchar('-');
	prt(x);
	if(ch) putchar(ch);
}

inline int query (int l, int r) {
	if(l == r) return 0;
	int ret;
	printf("? %d %d\n", l, r);
	fflush(stdout);
	scanf("%d", &ret);
	return ret;
}

void solve () {
	int n = read();
	if(n == 2) { printf("! %d\n", query(1, 2) == 1 ? 2 : 1); return; }
	int target = query(1, n);
	
	if(query(1, target) == target) { // maxinum is in [1, target)
		int l = 1, r = target - 1;
		int mid;
		
		while(l < r) {
			mid = l + r + 1 >> 1;
			if(query(mid, target) != target) r = mid - 1;
			else l = mid;
		}
		
		printf("! %d\n", l);
	}
	else { // maxinum is in (target, n]
		int l = target + 1, r = n;
		int mid;
		
		while(l < r) {
			mid = l + r >> 1;
			if(query(target, mid) == target) r = mid;
			else l = mid + 1;
		}
		
		printf("! %d\n", l);
	}
}

int main () {

	int T = 1;
//	T = read();
	while(T -- ) solve();

	return 0;
}