#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i=0; i<n; i++)
#define debug 0
#define YES cout << "Yes" << endl;
#define NO cout << "No" << endl;
using ll = long long;
using ld = long double;
const int mod = 998244353;
const int MOD = 1000000007;
const double pi = atan2(0, -1);
const int inf = 1 << 31 - 1;
const ll INF = 1LL << 63 - 1;
#include <time.h>
#include <chrono>

//vectorの中身を空白区切りで出力
template<typename T>
void printv(vector<T> v) {
	for (int i = 0; i < v.size(); i++) {
		cout << v[i];
		if (i < v.size() - 1) {
			cout << " ";
		}
	}
	cout << endl;
}

//vectorの中身を改行区切りで出力
template<typename T>
void print1(vector<T> v) {
	for (auto x : v) {
		cout << x << endl;
	}
}

//二次元配列を出力
template<typename T>
void printvv(vector<vector<T>> vv) {
	for (vector<T> v : vv) {
		printv(v);
	}
}

//vectorを降順にソート
template<typename T>
void rsort(vector<T>& v) {
	sort(v.begin(), v.end());
	reverse(v.begin(), v.end());
}

//昇順priority_queueを召喚
template<typename T>
struct rpriority_queue {
	priority_queue<T, vector<T>, greater<T>> pq;

	void push(T x) {
		pq.push(x);
	}

	void pop() {
		pq.pop();
	}

	T top() {
		return pq.top();
	}

	size_t size() {
		return pq.size();
	}

	bool empty() {
		return pq.empty();
	}
};

//mod mod下で逆元を算出する
//高速a^n計算(mod ver.)
ll power(ll a, ll n) {
	if (n == 0) {
		return 1;
	}
	else if (n % 2 == 0) {
		ll x = power(a, n / 2);
		x *= x;
		x %= mod;
		return x;
	}
	else {
		ll x = power(a, n - 1);
		x *= a;
		x %= mod;
		return x;
	}
}
//フェルマーの小定理を利用
ll modinv(ll p) {
	return power(p, mod - 2) % mod;
}

int main() {
	int N;
	cin >> N;
	queue<vector<int>> q;
	vector<int> ans;
	q.push({ 0,N-1,5 });
	int cnt = 0;
	while (ans.size() < 5 && cnt < 29) {
		cnt++;
		int mn = q.front()[0], mx = q.front()[1], num = q.front()[2];
		q.pop();
		int k = -1;

		int K = (mx - mn + 1)/2;
		cout << N << " " << K << endl;
		rep(i, K) {
			cout << mn + i << " ";
			k = max(k, mn + i);
		}
		cout << endl;

		int rep;
		cin >> rep;

		if (rep > 0) {
			if (k == mn) {
				rep(i, rep) {
					ans.push_back(mn);
				}
			}
			else {
				q.push({ mn, k, rep });
			}
		}
		if (rep < num) {
			if (k + 1 == mx) {
				rep(i, num - rep) {
					ans.push_back(mx);
				}
			}
			else {
				q.push({ k + 1, mx, num - rep });
			}
		}
	}

	int sum = 0;
	rep(i, 5) {
		sum += ans[i];
		if (ans[i] == 0) {
			sum += N;
		}
	}

	cout << 0 << " " << 1 << endl << sum << endl;
}