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 <iostream>
#include <vector>
#pragma GCC target ("avx,avx2,fma")
#pragma GCC optimization ("O8")
#pragma GCC optimization ("unroll-loops")
#define X first
#define Y second
#define debug(x) cerr << #x << ": " << x << endl;
#define endl '\n'
#define pb push_back
#define rep(i,l,r) for (int i=l; i<r; i++)
#define repr(i,r,l) for (int i=r; i>=l; i--)
using namespace std;
typedef long long  int ll;
typedef pair<ll,ll> pll;
const long long int N=1e3+20,mod = 1e9,inf=4e18,maxm = 2e5+20;
const long double eps = 0.0001;
void f(int l,int r){
    int m,x,y;
    cout << "? " << l << ' ' << r << endl;
    cin >> x;
    if (x != r && x != l){
        int y;
        cout << "? " << l << ' ' << x << endl;
        cin >> y;
        if (y == x){
            r = x-1;
            while(r-l > 1){
                m = (l+r)/2;
                if (m == x) continue;
                cout << "? " << m << ' ' << x << endl;
                cin >> y;
                if (y == x) l=m;
                else r = m;
            }
            if (l+1 == r){
                cout << "? " << l << ' ' << r << endl;
                cin >> y;
                if (y == l) l = r;
            }
            cout << "! " << l << endl;
            return;
        }
        l = x+1;
        while (r-l > 1){
            m = (l+r)/2;
            if (x == m) continue;
            cout << "? " << x << ' ' << m << endl;
            cin >> y;
            if (y == x) r = m;
            else (l = m);
        }
        if (l+1 == r){
            cout << "? " << l << ' ' << r << endl;
            cin >> y;
            if (y == r) r = l;
        }
        cout << "! " << r << endl;
        return;

    }
    if (x == l){
        l++;
        while (r-l > 1){
            m = (l+r)/2;
            if (x == m) continue;
            cout << "? " << x << ' ' << m << endl;
            cin >> y;
            if (y == x) r = m;
            else l = m;
        }
        if (l+1 == r){
            cout << "? " << l << ' ' << r << endl;
            cin >> y;
            if (y == r) r = l;
        }
        cout << "! " << r << endl;
        return;
    }
    r = x-1;
    while (r-l > 1){
        m = (l+r)/2;
        if (m == x) continue;
        cout << "? " << m << ' ' << x << endl;
        cin >> y;
        if (y == x) l=m;
        else r = m;
    }
    if (l+1 == r){
        cout << "? " << l << ' ' << r << endl;
        cin >> y;
        if (y == l) l = r;
    }
    cout << "! " << l << endl;
}
int main(){
    //ios :: sync_with_stdio(0); cin.tie(0);cout.tie(0);
    int n;
    cin >> n;
    f(1,n);
    return 0;
}