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 F first
#define S second
#define MP make_pair
#define PB push_back

using namespace std;

typedef long long ll;
typedef long double ld;

const int max_n = 1011, inf = 1000111222;
const int mod = 1000000007;

int n, r, arr[max_n];
bool debug = false;

set<pair<int, int>> get_s(int l, int r) {
  set<pair<int, int>> s;
  for (int i = l; i <= r; ++i) {
    s.insert({arr[i], i});
  }
  return s;
}

void rev(int& ind) {
  if (!r) return;
  ind = n - ind + 1;
}

bool ans(int ans) {
  if (!debug) {
    cout << "! " << ans << endl;
  } else {
    set<pair<int, int>> s = get_s(1, n);
    return (*(--s.end())).second == ans;
  }
  return true;
}

int debug_query(int l, int r) {
  set<pair<int, int>> s = get_s(l, r);
  return (--(--s.end()))->second;
}

int query(int l, int r) {
  rev(l);
  rev(r);
  assert(l < r);
  int ind;
  if (!debug) {
    cout << "? " << l << " " << r << endl;
    cin >> ind;
  } else {
    ind = debug_query(l, r);
    // cout << ind << endl;
  }
  rev(ind);
  return ind;
}

int bin_p(int sl, int sr, int l, int r, int is_l, int pos) {
  while (l + 1 < r) {
    int mid = (l + r) / 2;
    if (is_l) {
      if (query(sl, mid - 1) == pos) {
        r = mid;
      } else {
        l = mid;
      }
    } else {
      if (query(mid, sr) == pos) {
        l = mid;
      } else {
        r = mid;
      }
    }
  }
  return l;
}

void solve() {
  if (debug) {
    n = rand() % 5 + 2;
    for (int i = 0; i < n; ++i) {
      arr[i] = i + 1;
    }
    random_shuffle(arr, arr + n);
    // cout << n << endl;
    // for (int i = 0; i < n; ++i) cout << arr[i] << " "; cout << endl;
  } else {
    cin >> n;
  }
  int pos = query(1, n);
  int l, r;
  l = 1, r = n;
  int res = -1;
  if (pos == n) {
    res = bin_p(1, n, 1, n, 0, n);
  } else if (pos == 1) {
    res = bin_p(1, n, 2, n + 1, 1, 1);
  } else {
    int pos2 = query(1, pos);
    if (pos2 == pos) {
      res = bin_p(1, n, 1, pos, 0, pos);
    } else {
      res = bin_p(1, n, pos + 1, n + 1, 1, pos);
    }
  }
  if (!ans(res)) {
    cout << "FAIL" << endl;
    exit(1);
  } else if (debug) {
    cout << "OK\n";
  }
}


int main() {
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    do {
      solve();
    } while (debug);
    return 0;
}