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 int long long
#define endl "\n"
#define MOD 1000000007
#define INF LLONG_MAX
#define MINF LLONG_MIN
#define FASTIO ios_base::sync_with_stdio(false);cin.tie(NULL); cout.tie(NULL);
#define vi vector<int>
#define vvi vector<vector<int>>
#define pii pair<int, int>
#define mii map<int, int>
#define si set<int>

int pow(int x, int y, int mod) {
    int temp;
    if (y == 0)
        return 1;

    temp = (pow(x, y / 2, mod)) % mod;

    if (y % 2 == 0)
        return (temp * temp) % mod;
    else
        return (((x * temp) % mod) * temp) % mod;
}

map<pair<int, int>, int> mp;

int ask(int l, int r) {

    if(mp.count({l, r}))
        return mp[{l, r}];

    cout << "? "<< l << ' '<< r << endl;
    cout << flush;

    int ind; cin >> ind;
    return mp[{l, r}] = ind;
}

void answer(int i) {
    cout << "! "<< i << endl;
    cout << flush;
}

void solve(int tno) {

    int n; cin >> n;

    int scnd = ask(1, n);


    int lf ;
    
    if(scnd != 1)
        lf = ask(1, scnd);
    else
        lf = 2;

    int l, r, m;

    if(lf == scnd) {
        l = 1, r = scnd - 1;
        while(r - l > 1) {

            m = l + (r - l) / 2;

            int i = ask(m, scnd);

            if(i == scnd) {
                l = m;
            }
            else
                r = m;
        }

        if(ask(r, scnd) == scnd)
            answer(r);
        else
            answer(l);
    }

    else {
        
        l = scnd + 1, r = n;
        while(r - l > 1) {

            m = l + (r - l) / 2;

            int i = ask(scnd, m);

            if(i == scnd) {
                r = m;
            }
            else
                l = m;
        }

        if(ask(scnd, l) == scnd)
            answer(l);
        else
            answer(r);
    }
}

int32_t main() {

    FASTIO

// #ifndef ONLINE_JUDGE
//     freopen("input.txt", "r", stdin);
//     freopen("output.txt", "w", stdout);
// #endif

    int t;
    // cin >> t;
    t = 1;
    for (int i = 1; i <= t; i ++)
        solve(i);
}