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 PB emplace_back
#define MP make_pair
#define all(x) (x).begin(), (x).end()

using namespace std;
typedef long long LL;
int t, n;

int query(int l, int r)
{
    int res = 0;
    cout << "? " << l << " " << r << "\n";
    cout.flush();
    cin >> res;
    return res;
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);

    cin >> n;
    int l = 1, r = n;
    int pos = query(l, r);

    if (l != pos)
    {
        if (query(l, pos) == pos)  r = pos;
        else  l = pos;
	}

    if (l == pos)  // 答案在右边
    {
    	// 满足条件的最小下标
        /*while (l < r)
        {
            if (pos + 1 == r)  break;
            int m = l + r >> 1;
            if (query(pos, m) == pos)  r = m;
            else  l = m + 1;
		}*/
        while (l < r - 1)
        {
            int m = l + r >> 1;
            if (query(pos, m) == pos)  r = m;
            else  l = m;
        }
		cout << "! " << r << endl;
    }
    else  // 答案在左边
    {
    	// 满足条件的最大下标
        while (l < r)
        {
            if (pos - 1 == l)  break;
            int m = (l + r + 1) >> 1;
            if (query(m, pos) == pos)  l = m;
            else  r = m - 1;
		}
		cout << "! " << l << endl;
    }

    return 0;
}