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;

typedef long long ll;

int q(int l, int r){
    cout << "? " << l << " " << r << endl;
    int idx; cin >> idx;
    return idx;
}
void solve(){
    int n;
    cin >> n;
    int idx = q(1, n);
    int L = -1, R = -1;
    if(1 < idx) L = q(1, idx);
    if(n > idx) R = q(idx, n);
    if(R == idx){ // 最大值在右
        int ans = n, l = idx + 1, r = n, mid;
        while(l <= r){
            mid = l + r >> 1;
            int t = q(idx, mid);
            if(t == idx){
                ans = min(ans, mid);
                r = mid - 1;
            }else{
                l = mid + 1;
            }
        }
        cout << "! " << ans << endl;
    }else{
        int ans = 1, l = 1, r = idx - 1, mid;
        while(l <= r){
            mid = l + r >> 1;
            int t = q(mid, idx);
            if(t == idx){
                ans = max(ans, mid);
                l = mid + 1;
            }else{
                r = mid - 1;
            }
        }
        cout << "! " << ans << endl;
    }
}
// 5 1 2 3 4
int main(){
    // #define LOCAL
#ifdef LOCAL
    freopen("in.in", "r", stdin);
    freopen("out.out", "w", stdout);
#endif  
    int _ = 1;
    // cin >> _;
    for(int i = 1; i <= _; ++i) solve();
}