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;
typedef long double ld;
typedef vector<int> vi;
#define eb emplace_back
#define all(x) begin(x), end(x)
#define FOR(i, a, b) for(int i=(a); i<=(b); i++)
#define ROF(i, a, b) for(int i=(a); i>=(b); i--)
#define chkmax(x, a) x = max(x, a)
#define chkmin(x, a) x = min(x, a)

int query(int l, int r) {
  if (l == r) return -1;
  cout << "? " << l << " " << r << endl;
  cout.flush();
  int x;
  cin >> x;
  return x;
}

void report(int x) {
  cout << "! " << x << endl;
  cout.flush();
}

void solve(){
  int n; cin >> n;
  int k = query(1, n);
  if (k != 1 && query(1, k) == k) {
    int lo = 1, hi = k;
    while (lo < hi) {
      int m = lo + hi >> 1;
      if (query(m, k) != k) {
        hi = m;
      }else{
        lo = m + 1;
      }
    }
    report(lo-1);
  }else{
    int lo = k+1, hi = n;
    while (lo < hi) {
      int m = lo + hi >> 1;
      if (query(k, m) == k) hi = m;
      else lo = m + 1;
    }
    report(hi);
  }
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(NULL); cout.tie(NULL);

  int T = 1;
  // cin >> T;
  while (T--) solve();
  return 0;
}