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>
using namespace std;
#define ll long long
#define ull unsigned long long
#define db double
#define pii pair<int,int>
#define pli pair<ll,int>
#define pil pair<int,ll>
#define pll pair<ll,ll>
#define ti3 tuple<int,int,int>
#define mat vector<vector<int>>
const int inf = 1 << 30;
const ll linf = 1LL << 62;
const db EPS = 1e-7;
template<class T> void chmin(T& x, T y){if(x > y) x = y;}
template<class T> void chmax(T& x, T y){if(x < y) x = y;}

void solve(){
    int n;
    cin >> n;

    int l = 1, r = n, sec;
    cout << "?" << " " << l << " " << r << endl;
    cout.flush();
    cin >> sec;

    if(l != sec){
        int nxt;
        cout << "?" << " " << l << " " << sec << endl;
        cout.flush();
        cin >> nxt;
        if(nxt == sec){
            l = sec;
            for(int i = 16; i >= 0; i--){
                if(l - (1 << i) < 1) continue;
                cout << "? " << l - (1 << i) << " " << sec << endl;
                cout.flush();
                cin >> nxt;
                if(nxt != sec) l -= 1 << i;
            }
            cout << "! " << l - 1 << endl;
            cout.flush();
            return;
        }
    }
    if(r != sec){
        int nxt;
        r = sec;
        for(int i = 16; i >= 0; i--){
            if(r + (1 << i) > n) continue;
            cout << "? " << sec << " " << r + (1 << i) << endl;
            cout.flush();
            cin >> nxt;
            if(nxt != sec) r += 1 << i;
        }
        cout << "! " << r + 1 << endl;
        cout.flush();
        return;
    }
}

int main(){
    solve();
    return 0;
}