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 <tuple>
#include <random>
#include <cmath>
#include <numeric>
#include <cassert>
#include <iomanip>
#include <algorithm>
#include <set>

#define sz(a) (int)(a).size()
#define all(a) begin(a), end(a)

using namespace std;

using ll = long long;
using ull = unsigned long long;
using ld = long double;

ll dist(int x1, int y1, int x2, int y2) {
    return abs(x1 - x2) + abs(y1 - y2);
}

int main() {
#ifdef LOCAL
//    freopen("input.txt", "r", stdin);
#endif

    int n;
    cin >> n;

    cout << "? " << 1 << " " << n << endl;
    int m;
    cin >> m;

    bool is_lefter = false;

    if (m > 1) {
        cout << "? " << 1 << " " << m << endl;
        int ans;
        cin >> ans;
        if (ans == m) {
            is_lefter = true;
        }
    }

    if (is_lefter) {
        int l = 1;
        int r = m;
        while (r - l > 1) {
            int q = (l + r) / 2;
            cout << "? " << q << " " << m << endl;
            int ans;
            cin >> ans;
            if (ans == m) {
                l = q;
            } else {
                r = q;
            }
        }
        cout << "! " << l << endl;
    } else {
        int l = m;
        int r = n;
        while (r - l > 1) {
            int q = (l + r) / 2;
            cout << "? " << m << " " << q << endl;
            int ans;
            cin >> ans;
            if (ans == m) {
                r = q;
            } else {
                l = q;
            }
        }
        cout << "! " << r << endl;
    }

    return 0;
}