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:
//#pragma GCC optimize("ofast")
//#pragma GCC optimize("o3")
#include <bits/stdc++.h>
#define int long long
#define ll long long
#define fr(i,x,y) for(int i=x;i<y;i++)
#define rfr(i,x,y) for(int i=x;i>=y;i--)
using namespace std;
const int N=1e6,M=1e5,INF=2e18,MOD1=1e9+9,MOD2=1e9+7,L=20,P=11,Q=3e7;
int d[N];
bool used[N];
bool used2[N];
bool comp(pair<int,int> a,pair<int,int> b)
{
    return a.second<b.second;
}
vector<pair<int,int> > gr[N];
int uk1[N];
int uk2[N];
void solve()
{
    int n;
    cin>>n;
    cout<<"? "<<1<<' '<<n<<endl;
    int pos;
    cin>>pos;
    int l=1,r=pos;
    if(pos!=1)cout<<"? "<<l<<' '<<pos<<endl;
    int res;
    if(pos!=1)cin>>res;
    else res=0;
    if(res==pos)
    {
        l=0,r=pos;
        while(r-l>1)
        {
            //cout<<l<<' '<<r<<'\n';
            int mid=(r+l)/2;
            cout<<"? "<<mid<<' '<<pos<<endl;
            cin>>res;
            if(res==pos)l=mid;
            else r=mid;
        }
        cout<<"! "<<l<<endl;
    }
    else
    {
        l=pos,r=n+1;
        while(r-l>1)
        {
            int mid=(r+l)/2;
            cout<<"? "<<pos<<' '<<mid<<endl;
            cin>>res;
            if(res==pos)r=mid;
            else l=mid;
        }
        cout<<"! "<<r<<endl;
    }
}
signed main()
{
//freopen("021.dat","r",stdin);
//freopen("encryption.out","w",stdout);
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int t=1;
    //cin>>t;
    while(t--)solve();
}