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 inf          1e15
#define mod          1000000007
#define N            100001
#define ll           long long int
#define vint         vector<int>
#define vll          vector<ll>
#define vstr         vector<string>
#define vvint        vector<vector<int> >
#define vvll         vector<vector<ll> >
#define vint_pair    vector<pair<int,int> >
#define input(arr)   for(auto &x:arr) cin>>x;
#define all(x)       (x).begin(), (x).end()
#define rall(x)      (x).rbegin(), (x).rend()

int query(int l, int r){
    if(l==r)
        return -1;
    cout<<"? "<<l<<" "<<r<<endl;
    int idx;
    cin>>idx;
    return idx;
}

int main() {
    int t=1;
    //cin>>t;
    while(t--){
        int n;
        cin>>n;
        int mid,l=1,r=n;
        int scmx = query(l,r);
        int idx;
        if(scmx==query(l,scmx)){
            r = scmx-1;
            while(l<=r){
                mid = l + (r-l)/2;
                if(scmx == query(mid,scmx)){
                    idx = mid;
                    l = mid + 1;
                }
                else{
                    r = mid - 1;
                }
            }
        }
        else{
            l = scmx + 1;
            while(l<=r){
                mid = l + (r-l)/2;
                if(scmx == query(scmx,mid)){
                    idx = mid;
                    r = mid - 1;
                }
                else{
                    l = mid + 1;
                }
            }
        }
        cout<<"! "<<idx<<"\n";
    }
}