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 IOS                  \
    ios::sync_with_stdio(0); \
    cin.tie(0);              \
    cout.tie(0);
#define PB push_back
typedef long long ll;
#define mod 1000000007

map<pair<int, int>, int> qans;
int query(int l, int r)
{
    if (qans.count({l, r}))
    {
        return qans[{l, r}];
    }
    if (l == r)
    {
        return -1;
    }
    cout << "? " << l << " " << r << endl;
    int ans;
    cin >> ans;
    return qans[{l, r}] = ans;
}

void answer(int x)
{
    cout << "! " << x << endl;
    exit(0);
}

void solve()
{
    int n;
    cin >> n;
    int smx = query(1, n);
    int ans;
    if (smx == query(1, smx))
    {
        int L = 1, U = smx - 1;
        while (L <= U)
        {
            int m = (L + U) / 2;
            if (query(m, smx) == smx)
            {
                ans = m;
                L = m + 1;
            }
            else
            {
                U = m - 1;
            }
        }
    }
    else
    {
        int L = smx + 1, U = n;
        while (L <= U)
        {
            int m = (L + U) / 2;
            if (query(smx, m) == smx)
            {
                ans = m;
                U = m - 1;
            }
            else
            {
                L = m + 1;
            }
        }
    }
    answer(ans);
}

int main()
{
    IOS;
    solve();
    return 0;
}