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>
 
#define FOR(I,A,B)   for(int I= (A); I<(B); ++I)
#define REP(I,N)     FOR(I,0,N)
#define ALL(A)       (A).begin(), (A).end()
#define SZ(A)        int((A).size())
#define PB           push_back
#define MP           make_pair
#define FST(A)       (A).first
#define SEC(A)       (A).second
#define DEBUG 0
 
using namespace std;
 
typedef long long int ll;
typedef unsigned long long int ulld;
typedef pair<int,int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
typedef vector<bool> vb;

int main () {
  int n;
  cin >> n;
  // Get index of s_h
  cout << "? 1 " << n << endl;
  int s_h;
  cin >> s_h;
  // is the max on the left?
  int q;
  if (s_h != 1) {
    cout << "? 1 " << s_h << endl;
    cin >> q;
  }
  if (s_h != 1 && q == s_h) {
    // on the left. query will always be ? x s_h
    int l = 1, r = s_h;
    while (l < r) {
      if (r - l == 1) {
        break;
      }
      int index = (l + r) / 2;
      cout << "? " << index << " " << s_h << endl;
      cin >> q;
      if (q == s_h) {
        // The max is in this range still... go lower.
        l = index;
      } else {
        // The max is not in this range...
        r = index - 1;
      }
    }
    if (l == r) {
      cout << "! " << l << endl;
    } else {
      if (r == s_h) {
        cout << "! " << l << endl;
      } else {
        // try r.
        cout << "? " << r << " " << s_h << endl;
        cin >> q;
        if (q == s_h) {
          cout << "! " << r << endl;
        } else {
          cout << "! " << l << endl;
        }
      }
    }
  } else {
    // on the right. query will always be ? s_h x
    int l = s_h, r = n;
    while (l < r) {
      if (r - l == 1) {
        break;
      }
      int index = (l + r) / 2;
      cout << "? " << s_h << " " << index << endl;
      cin >> q;
      if (q == s_h) {
        // The max is in this range still... go lower.
        r = index;
      } else {
        // The max is not in this range...
        l = index + 1;
      }
    }
    if (l == r) {
      cout << "! " << l << endl;
    } else {
      if (l == s_h) {
        cout << "! " << r << endl;
      } else {
        // try r.
        cout << "? " << s_h << " " << l << endl;
        cin >> q;
        if (q == s_h) {
          cout << "! " << l << endl;
        } else {
          cout << "! " << r << endl;
        }
      }
    }
  }
}