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;
using ll = long long;

int n;
map<pair<int,int>,int> ma;
int query(int s, int e){
    int result;
    if(ma.find({s,e})==ma.end()){
        cout << "? " << s+1 << " " << e+1 << '\n';
        cin >> result;
        result--;
        ma.insert({{s,e},result});
    } else result = ma.find({s,e})->second;
    return result;
}
void answer(int n){
    cout << "! " << n+1;
}
int param1(int s, int e, int second){
    if(e-s==1) return e;
    int mid = (s+e)/2;
    bool result = true;
    if(second+mid<n){
        int max_val = query(second,second+mid);
        if(max_val != second) result = false;
    }
    if(result) return param1(s,mid,second);
    else return param1(mid,e,second);
}
int param2(int s, int e, int second){
    if(e-s==1) return e;
    int mid = (s+e)/2;
    bool result = true;
    if(second-mid>=0){
        int max_val = query(second-mid,second);
        if(max_val != second) result = false;
    }
    if(result) return param2(s,mid,second);
    else return param2(mid,e,second);
}
int main(void){
    // ios::sync_with_stdio(0);
    // cin.tie(0);

    cin >> n;
    int s = 0, e = n-1;
    int chk = -1;
    int val = query(s,e);
    if(s<val) chk = query(s,val);
    if(chk!=val){
        answer(val+param1(0,n,val));
    } else {
        answer(val-param2(0,n,val));
    }
    return 0;
}