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.
#include <unordered_set>
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <iterator>
#include <numeric>
#include <sstream>
#include <fstream>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cstdio>
#include <vector>
#include <bitset>
#include <cmath>
#include <queue>
#include <tuple>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
using namespace std;

typedef long long Int;
#define rep(i,a,n) for (Int i=a;i<n;i++)
#define per(i,a,n) for (Int i=n-1;i>=a;i--)
#define endl '\n' // Normal `cout << endl` flushes the output every time wich hit performance badly
#define INF 9223372036854775807LL
#define NINF -9223372036854775807LL
#define printArray(arr) for(auto x: arr) { cout<<x<<"\t"; }
#define printMatrix(mat) for(auto x: mat) { cout<<" "; printArray(x); cout<<endl; }
#define printMap(mmap) for(auto p: mmap) { cout<< p.first<<": "<<p.second<<endl; }
#define MIN(a,b) ((a)<(b)?(a):(b))
#define MAX(a,b) ((a)>(b)?(a):(b))
#define CJCOAX 1

//int read(int i) {
//    cout<<"? "<<i<<endl;
//    cout.flush();
//    int cur;
//    cin>>cur;
//    return cur;
//}
//
//#if CJCOAX
//int main(){
//    ios_base::sync_with_stdio(false); cin.tie(NULL); // Fast input
//
//    int n;
//    cin>>n;
//    vector<int> arr(n+2,-1);
//    int low = 1, high = n, ans = 0;
//    arr[0] = 1000000;
//    arr[n+1] = 1000000;
//    while (low < high) {
//        int mid = low + (high-low)/2;
//
//        if (mid-1 >= 0 && arr[mid-1] == -1) {
//            arr[mid-1] = read(mid-1);
//        }
//        if (arr[mid] == -1) {
//            arr[mid] = read(mid);
//        }
//        if (mid+1 <= n && arr[mid+1] == -1) {
//            arr[mid+1] = read(mid);
//        }
//        if (arr[mid] < arr[mid-1] && arr[mid] < arr[mid+1]) {
//            cout<<"! "<<mid<<endl;
//            cout.flush();
//            break;
//        } else if (arr[mid] > arr[mid-1]) {
//            high = mid;
//        } else {
//            low = mid;
//        }
//    }
//    return 0;
//}

const int MAXN = 100010;

int n;
int a[MAXN];
 
void query(int x)
{
    if (1 <= x && x <= n)
    {
        printf("? %d\n", x);
        fflush(stdout);
        scanf("%d", &a[x]);
    }
}
 
int main()
{
    scanf("%d", &n);
    a[0] = a[n + 1] = n + 1;
    int L = 1, R = n;
    while (L < R)
    {
        int m = (L + R) / 2;
        query(m);
        query(m + 1);
        if (a[m] < a[m + 1])
            R = m;
        else
            L = m + 1;
    }
    printf("! %d\n", L);
    fflush(stdout);
    return 0;
}

