#include <bits/stdc++.h>
#define int long long
using namespace std;
const long long N = 2e5 + 5;
int a[N];
int n;
bool chk(int l, int r)
{
if (r >= n)
return 0;
else if (l < 0)
return 0;
bool ans;
cout << "? " << l + 1 << " " << r + 1 << endl;
cin >> ans;
return ans;
}
int bin(int L, int R)
{
if (!chk(L, R))
return 0;
int l = 0, r = min(n - R, L) + 2;
while (r - l > 1)
{
int mid = (r + l) / 2;
chk(L - mid, R + mid) ? (l = mid) : (r = mid);
}
return R + l + 1 - (L - l);
}
void solve()
{
cin >> n;
int ans = 1;
for (int i = 0; i < n; i++)
{
// cout << " i = " << i << endl;
// single character is middle
// at least ans of them should make a palindrome
// cout << " odd ------------------" << endl;
// if (previous palindrom was even length) lets say 4 length, now we need a 5 length one, so 2 + middle + 2
if (ans % 2 == 0)
{
ans = max(ans, bin(i - ans / 2, i + ans / 2));
}
// if (previous palindrom was odd length) lets say 5 length, now we need a 7 length one, so 3 + middle + 3
else
{
ans = max(ans, bin(i - (ans + 1) / 2, i + (ans + 1) / 2));
}
// cout << " even ------------------" << endl;
// two in middle
// at least ans of them should make a palindrome
// if (previous palindrom was even length) lets say 8 length, now we need a 10 length one, so 4 + middle +(middle+1) + 4
if (ans % 2 == 0)
ans = max(ans, bin(i - ans / 2, i + ans / 2 + 1));
// if (previous palindrom was odd length) lets say 1 length, now we need a 2 length one, so 0 + middle + (middle+1) + 00
else
ans = max(ans, bin(i - ans / 2, i + ans / 2 + 1));
}
cout << "! "<< ans << endl;
}
signed main()
{
// ios_base::sync_with_stdio(0);
// cin.tie(NULL);
// cout.tie(NULL);
int t = 1;
// cin >> t;
for (int i = 1; i <= t; i++)
{
// cout << "Case #" << i << ':' << ' ';
solve();
cout << endl;
}
return 0;
}
