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:
/*
    Problem: Guessing the Greatest (easy version)
    Link: https://codeforces.com/contest/1486/problem/C1
*/

#include <bits/stdc++.h>
using namespace std;

#define debug(T) cout << ">> " << #T << " = " << (T) << endl;

typedef long long ll;
typedef unsigned long long ull;

// void preprocess() {}

int query(int l, int r) {
    if (l == r) {
        return -1;
    }

    cout << "? " << (l+1) << " " << (r+1) << endl;
    int idx;
    cin >> idx;
    return (idx-1);
}

int search(int l, int r, int smx_idx) { 
    if (l == r) {
        return l;
    }

    int m = (l+r)/2;

    if (smx_idx == -1) {
        smx_idx = query(l, r);
        if (smx_idx == query(l, smx_idx)) {
            return search(l, smx_idx, smx_idx);
        }
        else {
            return search(smx_idx, r, smx_idx);
        }
    }

    if (smx_idx <= m) {
        if (smx_idx == query(smx_idx, m)) {
            return search(l, m, smx_idx);
        }
        else {
            return search(m+1, r, smx_idx);
        }
    }
    else {
        if (smx_idx == query(m+1, smx_idx)) {
            return search(m+1, r, smx_idx);
        }
        else {
            return search(l, m, smx_idx);
        }
    }
}

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

    int ans = search(0, n-1, -1);
    cout << "! " << (ans+1) << endl;
}

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

    int test_case = 1;
    // cin >> test_case;

    // preprocess();
    while (test_case--) {
        solve();
        // cout << endl;
    }

    return 0;
}