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>
#include<iostream>
#include<iomanip>
#include<string>
#include<vector>
#include<algorithm>
#include<utility>
#include<tuple>
#include<map>
#include<queue>
#include<deque>
#include<set>
#include<stack>
#include<numeric>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
using namespace std;

#define forr(i,n) for( int i = 0 ; i < n ; i++)
#define forr1(i,n) for( int i = 1 ; i<= n ; i++)
#define pb push_back
#define int long long int
#define endl "\n"
typedef long long ll;
int  n , m ;
int mad = 1e9 + 7;


int32_t  main()
{

// #ifndef ONLINE_JUDGE
//     freopen("input.txt" , "r" , stdin);
//     freopen("output.txt" , "w" , stdout);
// #endif


    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    // string ss = "RLDU";
    // int cx[]  = {1, -1, 0 , 0};
    // int cy[] = {0, 0, -1, 1};


    int n ;
    cin >> n;

    int l  = 1 , r = n ;
    int prev ;
    cout << '?' << " " << l << " " << r << "\n";
    cout.flush();
    cin >> prev ;
    int curr;
    // r = prev;
    if ( n == 2)
    {
        if (prev == 1)
            cout << '!' << " " << 2 << "\n";
        else
            cout << '!' << " " << 1 << "\n";
        return 0 ;
    }

    if ( n > 2)
    {
    	int z ;
    	  if(prev >= r) z = -1;
    	  else
    	  {
        cout << '?' << " " << prev << " " << r << "\n";
        cout.flush(); 
        cin >> z;
         }
        if (prev == 1 || z == prev )
        {
            int   lfix = prev ;
            l = prev ; r = n ;
            while ( r - l > 1)
            {
                int mid = (l + r)/2;
                if(lfix  >= mid)
                	curr = -1 ;
                else
                {
                cout << '?' << " " << lfix << " " << mid << "\n";
                cout.flush();
                cin >> curr;
                }
                if (curr == prev)
                {
                    r = mid ;
                }
                else
                {
                    l = mid;
                }
            }
          
            cout << '!' << " " << r << endl;
        }
        else
        {

            int  rfix = prev ;
            r = prev ; l = 1;
            while ( r - l > 1)
            {
                int mid = (l + r) /2;
                if(rfix <= mid)
                	curr = -1;
                else
                {
                cout << '?' << " " << mid << " " << rfix << "\n";
                cout.flush();
                cin >> curr;
            }

                if (curr == prev)
                {
                    l = mid ;
                }
                else
                {
                    r = mid;
                }

            }

            cout << '!' << " " << l << endl;

        }
    }





}