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;
typedef long long ll;
const int N = 1e5+7;
const int MOD = 1e9+7;
int query(int x,int y){
    if(x==y){
        return -1;
    }
    cout<<"? "<<x<<" "<<y<<endl;
    int t;
    cin>>t;
    return t;
}
void solve(int x){
    cout<<"! "<<x<<endl;
    exit(0);
}
int n,a[N],S,L,R;
int vis[N];
void solveL(int le,int ri){
    int l=le,r=ri,ans=0;
    while(l<=r){
        int mid=l+r>>1;
        int t=query(1,mid);
        if(t==-1){
            solve(S==1?2:1);
        }
        if(t==S){
            r=mid-1;
            ans=mid;
        }else{
            l=mid+1;
        }
    }
    solve(ans);
}
void solveR(int le,int ri){
    int l=le,r=ri,ans=0;
    while(l<=r){
        int mid=l+r>>1;
        int t=query(mid,n);
        if(t==-1){
            solve(S==n?n-1:n);
        }
        if(t==S){
            l=mid+1;
            ans=mid;
        }else{
            r=mid-1;
        }
    }
    solve(ans);
}
int main(){
    cin>>n;
    if(n==2){
        int t=query(1,2);
        solve(3-t);
    }
    int l=1,r=n;
    int mid=l+r>>1;
    S=query(1,n);
    if(S==1){
        solveL(1,n);
    }else if(S==n){
        solveR(1,n);
    }else{
        L=query(1,S);
        if(L==S)solveR(1,S);
        else solveL(S,n);
    }
    
}