This is an interactive problem.

Homer likes arrays a lot and he wants to play a game with you. 

Homer has hidden from you a permutation a_1, a_2, ..., a_n of integers 1 to n. You are asked to find any index k (1 ≤ k ≤ n) which is a local minimum. 

For an array a_1, a_2, ..., a_n, an index i (1 ≤ i ≤ n) is said to be a local minimum if a_i < min\\{a_{i-1},a_{i+1}\}, where a_0 = a_{n+1} = +∞. An array is said to be a permutation of integers 1 to n, if it contains all integers from 1 to n exactly once.

Initially, you are only given the value of n without any other information about this permutation.

At each interactive step, you are allowed to choose any i (1 ≤ i ≤ n) and make a query with it. As a response, you will be given the value of a_i. 

You are asked to find any index k which is a local minimum after at most 100 queries.

Interaction

You begin the interaction by reading an integer n (1≤ n ≤ 10^5) on a separate line.

To make a query on index i (1 ≤ i ≤ n), you should output "? i" in a separate line. Then read the value of a_i in a separate line. The number of the "?" queries is limited within 100.

When you find an index k (1 ≤ k ≤ n) which is a local minimum, output "! k" in a separate line and terminate your program. 

In case your query format is invalid, or you have made more than 100 "?" queries, you will receive Wrong Answer verdict. 

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.



Hack Format

The first line of the hack should contain a single integer n (1 ≤ n ≤ 10^5).

The second line should contain n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n).

Example

Input


5
3
2
1
4
5


Output


? 1
? 2
? 3
? 4
? 5
! 3

Note

In the example, the first line contains an integer 5 indicating that the length of the array is n = 5.

The example makes five "?" queries, after which we conclude that the array is a = [3,2,1,4,5] and k = 3 is local minimum.

// Problem: A. Searching Local Minimum
// Contest: Codeforces - Codeforces Round #700 (Div. 1)
// URL: https://codeforces.com/contest/1479/problem/A
// Memory Limit: 512 MB
// Time Limit: 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

#include <iostream>
#include <iosfwd>
#include <iomanip>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <cmath>
#include <cassert>
#include <cctype>
#include <climits>
#include <vector>
#include <bitset>
#include <set>
#include <queue>
#include <math.h>
#include <stack>
#include <map>
#include <deque>
#include <string>
#include <list>
#include <iterator>
#include <sstream>
#include <complex>
#include <fstream>
#include <functional>
#include <numeric>
#include <utility>
#include <algorithm>
#include <assert.h>
#include <unordered_map>
#pragma GCC optimize ("-O3")

using namespace std;

#define watch(x) cout << (#x) << " = " << (x) << endl
#define PI double(2 * acos(0.0))
#define LL long long
#define MOD 1000000007
#define all(x) (x).begin(), (x).end()
#define INF 1e15
int query(int a, int n){
      if(a == 0 || a == n + 1){
            return 1e7;
      }
      cout << "? " << a << endl;
      int p;
      cin >> p;
      return p;
}
int helper(int a, int n){
      int left = a - 1;
      int right = a + 1;
      int mid = a;
      int p = query(left, n);
      int q = query(mid, n);
      int r = query(right, n);
      if(p > q && q < r){
            return 0;
      }
      if(p < q){
            return 2;
      }
      else{
            return 1;
      }
}
int main() {
      int n;
      cin >> n;
      int start = 1;
      int end = n;
      int ans = -1;
      while(start <= end){
            int mid = (start + end) / 2;
            int p = helper(mid, n);
            if(p == 0){
                  ans = mid;
                  break;
            }
            if(p == 1){
                  start = mid + 1;
            }else{
                  end = mid - 1;
            }
      }
      cout << "! " << ans << endl;
}







































