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;

// clang-format off
// template {{{
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define f first
#define s second

#define sz(x) int((x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define trav(a, x) for (auto &a : x)

#define L1(u, ...) [&](auto &&u) { return __VA_ARGS__; }
#define L2(u, v, ...) [&](auto &&u, auto &&v) { return __VA_ARGS__; }

#define sort_by(x, y) sort(all(x), [&](const auto &l, const auto &r) { return y; })

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vs = vector<string>;

using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;

using vpii = vector<pii>;
using vpll = vector<pll>;
using vpdd = vector<pdd>;

template <typename T> void ckmin(T &a, const T &b) { a = min(a, b); }
template <typename T> void ckmax(T &a, const T &b) { a = max(a, b); }

mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

namespace __input {
    template <class T1, class T2> void re(pair<T1, T2> &p);
    template <class T> void re(vector<T> &a);
    template <class T, size_t SZ> void re(array<T, SZ> &a);

    template <class T> void re(T &x) { cin >> x; }
    void re(double &x) { string t; re(t); x = stod(t); }
    template <class Arg, class... Args> void re(Arg &first, Args &...rest) { re(first); re(rest...); }

    template <class T1, class T2> void re(pair<T1, T2> &p) { re(p.f, p.s); }
    template <class T> void re(vector<T> &a) { for (int i = 0; i < sz(a); i++) re(a[i]); }
    template <class T, size_t SZ> void re(array<T, SZ> &a) { for (int i = 0; i < SZ; i++) re(a[i]); }
}
using namespace __input;

namespace __output {
    template <typename T> struct is_outputtable { template <typename C> static constexpr decltype(declval<ostream &>() << declval<const C &>(), bool()) test(int) { return true; } template <typename C> static constexpr bool test(...) { return false; } static constexpr bool value = test<T>(int()); };
    template <class T, typename V = decltype(declval<const T &>().begin()), typename S = typename enable_if<!is_outputtable<T>::value, bool>::type> void pr(const T &x);

    template <class T, typename V = decltype(declval<ostream &>() << declval<const T &>())> void pr(const T &x) { cout << x; }
    template <class T1, class T2> void pr(const pair<T1, T2> &x);
    template <class Arg, class... Args> void pr(const Arg &first, const Args &...rest) { pr(first); pr(rest...); }

    template <class T, bool pretty = true> void prContain(const T &x) { if (pretty) pr("{"); bool fst = 1; for (const auto &a : x) pr(!fst ? pretty ? ", " : " " : "", a), fst = 0; if (pretty) pr("}"); }

    template <class T> void pc(const T &x) { prContain<T, false>(x); pr("\n"); }
    template <class T1, class T2> void pr(const pair<T1, T2> &x) { pr("{", x.f, ", ", x.s, "}"); }
    template <class T, typename V, typename S> void pr(const T &x) { prContain(x); }

    void ps() { pr("\n"); }
    template <class Arg> void ps(const Arg &first) { pr(first); ps(); }
    template <class Arg, class... Args> void ps(const Arg &first, const Args &...rest) { pr(first, " "); ps(rest...); }
}
using namespace __output;

#define __pn(x) pr(#x, " = ")
#ifdef ANAND_LOCAL
#define pd(...) pr("\033[1;31m"), __pn((__VA_ARGS__)), ps(__VA_ARGS__), pr("\033[0m"), cout << flush
#else
#define pd(...)
#endif

namespace __algorithm {
    template <typename T> void dedup(vector<T> &v) { sort(all(v)); v.erase(unique(all(v)), v.end()); }
    template <typename T> typename vector<T>::const_iterator find(const vector<T> &v, const T &x) { auto it = lower_bound(all(v), x); return it != v.end() && *it == x ? it : v.end(); }
    template <typename T> size_t index(const vector<T> &v, const T &x) { auto it = find(v, x); assert(it != v.end() && *it == x); return it - v.begin(); }
    template <typename I> struct _reversed_struct { I &v_; explicit _reversed_struct(I &v) : v_{v} {} typename I::reverse_iterator begin() const { return v_.rbegin(); } typename I::reverse_iterator end() const { return v_.rend(); } };
    template <typename I> _reversed_struct<I> reversed(I &v) { return _reversed_struct<I>(v); }
}
using namespace __algorithm;

namespace __io {
    void setIO() { ios_base::sync_with_stdio(0); cin.tie(0); cout << setprecision(15); }
}
using namespace __io;
// }}}
// Returns first i in [l, r] s.t. predicate(i) is true. Never evaluates r.
template <typename I, typename P> I binarysearch(const P &predicate, I l, I r) {
    l--;
    while (r - l > 1) {
        auto mid = l + (r - l) / 2;
        if (predicate(mid))
            r = mid;
        else
            l = mid;
    }
    return r;
}

// clang-format on

#ifdef ANAND_LOCAL
vi secret;
#endif

int query(int l, int r) {
    ps("?", l + 1, r);
    cout << flush;
    int result;

#ifdef ANAND_LOCAL
    auto max_it = max_element(secret.begin() + l, secret.begin() + r);
    int max_value = *max_it;
    *max_it = -1;
    result = max_element(secret.begin() + l, secret.begin() + r) - secret.begin();
    *max_it = max_value;
#else
    re(result);
    result--;
#endif

    return result;
}

void finish(int i) {
    ps("!", i + 1);

#ifdef ANAND_LOCAL
    assert(i == max_element(all(secret)) - secret.begin());
#endif
}

int main() {
    setIO();

    int n;
    re(n);

#ifdef ANAND_LOCAL
    secret.resize(n);
    re(secret);
#endif

    int second = query(0, n);
    bool go_right = second == 0 || query(0, second + 1) != second;

    auto test = [&](int count) {
        int l, r;
        if (go_right) {
            l = second;
            r = l + count;
        } else {
            r = second + 1;
            l = r - count;
        }

        if (r > n || l < 0)
            return true;

        return second == query(l, r);
    };

    int c = -1 + binarysearch(test, 2, n);

    finish(go_right ? second + c : second - c);

    return 0;
}