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:
//Faith over fear

#include<bits/stdc++.h>
using namespace std;
 
#define mp make_pair
#define mt make_tuple
#define fi first
#define se second
#define pb push_back
#define ll int
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define rep(i,n) for(i=0;i<n;i++)
#define forn(i, n) for (ll i = 0; i < (ll)(n); ++i)
#define for1(i, n) for (ll i = 1; i <= (ll)(n); ++i)
#define ford(i, n) for (ll i = (ll)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (ll i = (ll)(a); i <= (ll)(b); ++i)
#define fora(it,x) for(auto it:x)
#define PI 3.14159265
#define bhaag_basanti ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define endl "\n"
#define dbug(x) cout<<#x<<"="<<x<<endl;
 
typedef pair<ll, ll> pii;

typedef vector<ll> vi;
typedef vector<pii> vpi;
typedef vector<vi> vvi;
typedef long long i64;
typedef vector<i64> vi64;
typedef vector<vi64> vvi64;
typedef pair<i64, i64> pi64;
typedef long double ld;
 
template<class T> bool uin(T &a, T b) { return a > b ? (a = b, true) : false; }
template<class T> bool uax(T &a, T b) { return a < b ? (a = b, true) : false; }

ll ask(ll l,ll r){
    cout<<"? "<<l<<" "<<r<<endl;
    cout.flush();
    ll te;
    cin>>te;
    return(te);
}

int main(){
    ll n;
    cin>>n;

    ll smax=ask(1,n);
    if(smax==1){
            ll l=smax+1,r=n;
            while (l<r)
            {
                //cout<<l<<" "<<r<<endl;
                if((r-l)==1){
                    ll t1=ask(smax,l);
                    if(t1!=smax) l++;

                    break;
                }
                //[smax-n]
                ll m=(l+r+1)/2;
                ll te=ask(smax,m);
                if(te==smax) r=m;
                else l=m+1;
            }
            cout<<"! "<<l<<endl;
            cout.flush();
    }
    else{
        ll te=ask(1,smax);
        if(te==smax){
            ll l=1,r=smax-1;
           // cout<<"a";
            while (l<r){

                ll m=(l+r+1)/2;
                ll t1=ask(m,smax);
                if(t1==smax) l=m;
                else r=m-1;
            }
            cout<<"! "<<l<<endl;
            cout.flush();
        }
        else{
            ll l=smax+1,r=n;
            while (l<r)
            {
                //[smax-n]
                if((r-l)==1){
                    ll t1=ask(smax,l);
                    if(t1!=smax) l=r;
                    break;
                }

                ll m=(l+r+1)/2;
                ll t1=ask(smax,m);
                if(t1==smax) r=m;
                else l=m+1;
            }
            cout<<"! "<<l<<endl;
            cout.flush();
        }
    }
}