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;

#define ll long long
#define ld long double
#define F first
#define S second
#define all(v) v.begin(),v.end()
#define sz(v) (int)v.size()
#define precision cout << fixed << setprecision(15);

const int inf = 1e9;
const long long INF = 1e18;
const int mod = 1e9 + 7;
const int bit32 = log2(inf) + 3;
const int bit64 = log2(INF) + 3;

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
clock_t time_p = clock();

void ktj() {
  time_p = clock() - time_p;
  cerr << "Time elapsed : " << (float)(time_p)/CLOCKS_PER_SEC << "\n";
}

void pre() {}

void solve() {
  int n; cin >> n;

  auto ask = [=](int L, int R) {
    cout << "? " << L << ' ' << R << endl;
    int result; cin >> result;
    return result;
  };

  auto output = [=](int x) {
    cout << "! " << x << endl;
    exit(0);
  };

  int L = 1, R = n;
  int p = ask(L, R);
  if (n == 2) {
    output(3 - p);
  }

  auto maxPos = [=] (int L, int R, int pos) mutable {
    int ans = 1;
    while (L < R) {
      int M = (L + R) / 2;
      if (M == L or M == R) {
        output(L + R - ask(L, R));
      }
      p = ask(M, pos);
      if (p == pos) {
        L = M; ans = M;
      } else {
        R = M - 1;
      }
    }
    output(ans);
  };

  auto minPos = [=] (int L, int R, int pos) mutable {
    int ans = n;
    while (L < R) {
      int M = (L + R) / 2;
      if (M == L or M == R) {
        output(L + R - ask(L, R));
      }
      p = ask(pos, M);
      if (p == pos) {
        R = M; ans = M;
      } else {
        L = M + 1;
      }
    }
    output(ans);
  };

  if (p == L) {
    minPos(L, R, p);
  } else if (p == R) {
    maxPos(L, R, p);
  } else {
    int x = ask(L, p);
    if (x == p) {
      maxPos(L, p, p);
    } else {
      minPos(p, R, p);
    }
  }
}

#define GOOGLE 0
#define MULTIPLE_TC 0

int32_t main() {

  ios_base::sync_with_stdio(false);
  cin.tie(NULL);cout.tie(NULL); 

  pre();

  int t = 1, T;

  if (MULTIPLE_TC)
    cin >> t;

  for (T = 1; T <= t; T++) {
    if (GOOGLE)
      cout << "Case #" << T << ": ";
    solve();
  }

  ktj();
}