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>
#define de(x)   cerr << "  debug  " << #x << "  ==  " << x << endl;
using namespace std;
typedef long long  LL;
const LL mod = 1e9 + 7;
int mqpow(int x, int n) {
    int res = 1;
    while(n) {
        if(n & 1) res = 1ll * res * x % mod;
        n >>= 1;
        x = 1ll * x * x % mod;
    }
    return res;
}
const int maxn = 1e5 + 7;
int n;
void solve () {
    cin >> n;
    int l = 1, r = n, res = -1;
    bool  same = false;
    int be;
    cout << '?' << ' ' << 1 << ' ' << r << endl;
    fflush(stdout);
    cin >> be;
    if(n == 2) {
        if(be == 1) res = 2;
        else {
            res = 1;
        }
        cout << '!' << ' ' << res << endl;
        fflush(stdout);
        return ;
    }
    bool left = false;
    if(be != 1) {
        int to;
        cout << '?' << ' ' << 1 << ' ' << be << endl;
        fflush(stdout);
        cin >> to;
        left = (to != be);
    } else left = true;
    if(left) {
        l = be; r = n;
        while(l <= r) {
            if(l == r) {
                res = l;
                break;
            }
            int mid = l + r >> 1;
            int to;
            if(mid == l) {
               cout << '?' << ' ' << l << ' ' << r << endl;
               fflush(stdout); 
               cin >> to;
               if(to == l) {
                   res = r;
                   break;
               } else {
                   res = l;
                   break;
               }
            }
            cout << '?' << ' ' << be << ' ' << mid << endl;
            fflush(stdout);
            cin >> to;
            if(to == be) {
                r = mid;
            } else l = mid + 1;
        }
    } else {
        l = 1; r = be;
        while(l <= r) {
            if(l == r) {
                res = l;
                break;
            }
            int mid = l + r >> 1;
            int to;
            if(mid == l) {
               cout << '?' << ' ' << l << ' ' << r << endl;
               fflush(stdout); 
               cin >> to;
               if(to == l) {
                   res = r;
                   break;
               } else {
                   res = l;
                   break;
               }
            }
            cout << '?' << ' ' << mid << ' ' << be << endl;
            fflush(stdout);
            cin >> to;
            if(to == be) {
                l = mid;
            } else r = mid - 1;
        }
    }
    cout << '!' << ' ' << res << endl;
    fflush(stdout);
}
int main () {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int T = 1;
    // scanf("%d", &T);
    while(T --) {
        solve ();
    }
    #ifdef iyua
        system("pause");
    #endif 
    return 0;
}