#include <bits/stdc++.h>
using namespace std;
template <class P, class Q>
ostream& operator<<(ostream& stream, pair<P, Q> v) {
  stream << "(" << v.first << ',' << v.second << ")";
  return stream;
}
template <class T>
ostream& operator<<(ostream& stream, const vector<T> v) {
  stream << "[ ";
  for (int i = 0; i < (int)v.size(); i++) stream << v[i] << " ";
  stream << "]";
  return stream;
}
template <class T>
T smin(T& a, T b) {
  if (a > b) a = b;
  return a;
}
template <class T>
T smax(T& a, T b) {
  if (a < b) a = b;
  return a;
}
const long long M = 10004205361450474;
int ask(vector<long long> s) {
  int sz = s.size();
  cout << sz << ' ';
  for (int i = 0; i < sz; ++i) cout << s[i] << " \n"[i == sz - 1];
  fflush(stdout);
  int v = 0;
  cin >> v;
  if (v == -1) {
    exit(0);
  }
  return v;
}
const int N = 1e4 + 5;
long long dp[N][6];
long long f(int A, int B) {
  if (!B) return 0;
  if (dp[A][B] != -1) return dp[A][B];
  long long& ans = dp[A][B];
  ans = 0;
  const int W = A;
  for (int i = 0; i < W; ++i) {
    ans += f(A, B - 1) + 1;
    A = min(10000ll, A + f(A, B - 1) + 1);
  }
  ans += f(A, B - 1);
  return ans;
}
long long g(long long a, int b) { return f(min(10000ll, a), b); }
int main(void) {
  memset(dp, -1, sizeof(dp));
  const long long c1 = M - f(10000, 4);
  const long long c2 = 2046;
  const long long c3 = 6;
  long long l, r;
  int used = 5;
  --used;
  if (ask(vector<long long>{c1}) > 0) {
    l = c1 + 1;
    r = M;
    goto shit;
  }
  --used;
  if (ask(vector<long long>{c2}) > 0) {
    l = c2 + 1;
    r = c1 - 1;
    goto shit;
  }
  --used;
  if (ask(vector<long long>{c3}) > 0) {
    l = c3 + 1;
    r = c2 - 1;
  } else {
    l = 1;
    r = c3 - 1;
  }
shit:
  for (int cs = used - 1; cs >= 0; --cs) {
    vector<long long> s;
    for (long long i = l; i + g(i, cs) <= r && s.size() < min(10000ll, l);
         i += g(i, cs) + 1)
      s.push_back(i + g(i, cs));
    int v = ask(s);
    if (!v) {
      smin(r, s[0] - 1);
    } else if (v == s.size()) {
      smax(l, s.back() + 1);
    } else {
      smin(r, s[v] - 1);
      smax(l, s[v - 1] + 1);
    }
  }
  return 0;
}
