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;
int n;

map<pair<int,int>, int> mp;
int query(int l, int r){
    if(l == r)
        return -1;
    if(mp.find({l, r}) != mp.end())
        return mp[{l, r}];
    cout<<"? "<<l<<" "<<r<<endl;
    int x;
    cin>>x;
    return mp[{l,r}] = x;
}

int findPos(int l, int r){
    int x = query(1, n);
    int y = query(x, n);
    int answer;
    if(y == x){
        // visit right side
        int L = x + 1;
        int R = r;

        while(L <= R){
            int mid = (L+R)/2;
            if(query(x, mid) == x){
                answer = mid;
                R = mid - 1;
            }
            else {
                L = mid + 1;
            }
        }
    }
    else {
        // visit left side
        int L = l;
        int R = x - 1;

        while(L <= R){
            int mid = (L+R)/2;
            if(query(mid, x) == x){
                answer = mid;
                L = mid + 1;
            }
            else {
                R = mid - 1;
            }
        }

    }

    return answer;
}

int main(){
    n = 0;
    cin>>n;

    int ans = findPos(1, n);
    cout<<"! "<<ans<<endl;
    //exit(0);

    return 0;
}