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 maxn=2e5+100;
const ll inf=1e18;
int main()
{
    int T,n,l,r,pos,p,mid;
    //int flag,T,n,x;
    //freopen("text.in","r",stdin);
    scanf("%d",&n);
    printf("? %d %d\n",1,n);
    fflush(stdout);
    scanf("%d",&p);
    if (p!=n) {printf("? %d %d\n",p,n);
    fflush(stdout);
    scanf("%d",&pos);
    }else {
        pos=-1;
    }
    if (pos==p) {
    l=p; r=n;
    int ans=-1;
    int flag=0;
    while (l<=r) {
        mid=(l+r)/2;
        if (mid==p) {
            l=mid+1;
            continue ;
        }
        printf("? %d %d\n",p,mid);
        fflush(stdout);
        scanf("%d",&pos);
        if (pos!=p) {
            //ans=mid;
            l=mid+1;
        }else {
            flag=1;
            ans=mid;
            r=mid-1;
        }
    }
    //printf("wqe\n");
    //if (flag) {
        printf("! %d\n",ans);
        return 0;
    //}
    }
    int ans=-1;
    int flag=0;
    l=1; r=p;
    while (l<=r) {
        mid=(l+r)/2;
        if (mid==p) {
            r=mid-1;
            continue ;
        }
        printf("? %d %d\n",mid,p);
        fflush(stdout);
        scanf("%d",&pos);
        if (pos!=p) {
            //ans=mid;
            r=mid-1;
        }else {
            flag=1;
            ans=mid;
            l=mid+1;
        }
    }
    printf("! %d\n",ans);
    return 0;
}