Problem: There is a weighted tree with n nodes and n-1 edges. The nodes are conveniently labeled from 1 to n. The weights are positive integers at most 100. Define the distance between two nodes to be the sum of edges on the unique path between the nodes. You would like to find the diameter of the tree. Diameter is the maximum distance between a pair of nodes.

Unfortunately, the tree isn't given to you, but you can ask some questions about it. In one question, you can specify two nonempty disjoint sets of nodes p and q, and the judge will return the maximum distance between a node in p and a node in q. In the words, maximum distance between x and y, where x ∈ p and y ∈ q. After asking not more than 9 questions, you must report the maximum distance between any pair of nodes.

Interaction

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1 000). Description of the test cases follows.

The first line of each test case contains an integer n (2 ≤ n ≤ 100) — the number of nodes in the tree.

To ask a question, print "k_1\ k_2\ a_1\ a_2\ …\ a_{k_1}\ b_1\ b_2\ …\ b_{k_2}" (k_1, k_2 ≥ 1 and k_1+k_2 ≤ n). These two sets must be nonempty and disjoint. The judge will respond with a single integer max_{p,q} dist(a_p, b_q). If you ever get a result of -1 (because you printed an invalid query), exit immediately to avoid getting other verdicts.

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.



When you are ready to answer, print "-1\ d", where d is the maximum shortest distance over all pairs of nodes.

You can only ask at most 9 questions per test case.

Hack Format

To hack, use the following format. Note that you can only hack with one test case.

The first line should contain a single integer t (t=1).

The second line should contain a single integer n (2 ≤ n ≤ 100).

Each of the next n-1 lines should contain three integers a_i, b_i, c_i (1≤ a_i, b_i≤ n, 1 ≤ c_i ≤ 100). This denotes an undirected edge between nodes a_i and b_i with weight c_i. These edges must form a tree.

Example

Input


2
5
9
6
10
9
10
2
99


Output


1 4 1 2 3 4 5
1 4 2 3 4 5 1
1 4 3 4 5 1 2
1 4 4 5 1 2 3
1 4 5 1 2 3 4
-1 10
1 1 1 2
-1 99

Note

In the first example, the first tree looks as follows: <image>

In the first question, we have p = {1}, and q = {2, 3, 4, 5}. The maximum distance between a node in p and a node in q is 9 (the distance between nodes 1 and 5).

The second tree is a tree with two nodes with an edge with weight 99 between them.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:36777216")
template <class T>
inline T &RD(T &);
template <class T>
inline void OT(const T &);
inline long long RD() {
  long long x;
  return RD(x);
}
inline double &RF(double &);
inline double RF() {
  double x;
  return RF(x);
}
inline char *RS(char *s);
template <class T0, class T1>
inline T0 &RD(T0 &x0, T1 &x1) {
  RD(x0), RD(x1);
  return x0;
}
template <class T0, class T1, class T2>
inline T0 &RD(T0 &x0, T1 &x1, T2 &x2) {
  RD(x0), RD(x1), RD(x2);
  return x0;
}
template <class T0, class T1, class T2, class T3>
inline T0 &RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3) {
  RD(x0), RD(x1), RD(x2), RD(x3);
  return x0;
}
template <class T0, class T1, class T2, class T3, class T4>
inline T0 &RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4) {
  RD(x0), RD(x1), RD(x2), RD(x3), RD(x4);
  return x0;
}
template <class T0, class T1, class T2, class T3, class T4, class T5>
inline T0 &RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5) {
  RD(x0), RD(x1), RD(x2), RD(x3), RD(x4), RD(x5);
  return x0;
}
template <class T0, class T1, class T2, class T3, class T4, class T5, class T6>
inline T0 &RD(T0 &x0, T1 &x1, T2 &x2, T3 &x3, T4 &x4, T5 &x5, T6 &x6) {
  RD(x0), RD(x1), RD(x2), RD(x3), RD(x4), RD(x5), RD(x6);
  return x0;
}
template <class T0, class T1>
inline void OT(const T0 &x0, const T1 &x1) {
  OT(x0), OT(x1);
}
template <class T0, class T1, class T2>
inline void OT(const T0 &x0, const T1 &x1, const T2 &x2) {
  OT(x0), OT(x1), OT(x2);
}
template <class T0, class T1, class T2, class T3>
inline void OT(const T0 &x0, const T1 &x1, const T2 &x2, const T3 &x3) {
  OT(x0), OT(x1), OT(x2), OT(x3);
}
template <class T0, class T1, class T2, class T3, class T4>
inline void OT(const T0 &x0, const T1 &x1, const T2 &x2, const T3 &x3,
               const T4 &x4) {
  OT(x0), OT(x1), OT(x2), OT(x3), OT(x4);
}
template <class T0, class T1, class T2, class T3, class T4, class T5>
inline void OT(const T0 &x0, const T1 &x1, const T2 &x2, const T3 &x3,
               const T4 &x4, const T5 &x5) {
  OT(x0), OT(x1), OT(x2), OT(x3), OT(x4), OT(x5);
}
template <class T0, class T1, class T2, class T3, class T4, class T5, class T6>
inline void OT(const T0 &x0, const T1 &x1, const T2 &x2, const T3 &x3,
               const T4 &x4, const T5 &x5, const T6 &x6) {
  OT(x0), OT(x1), OT(x2), OT(x3), OT(x4), OT(x5), OT(x6);
}
inline double &RF(double &a, double &b) {
  RF(a), RF(b);
  return a;
}
inline double &RF(double &a, double &b, double &c) {
  RF(a), RF(b), RF(c);
  return a;
}
inline double &RF(double &a, double &b, double &c, double &d) {
  RF(a), RF(b), RF(c), RF(d);
  return a;
}
inline double &RF(double &a, double &b, double &c, double &d, double &e) {
  RF(a), RF(b), RF(c), RF(d), RF(e);
  return a;
}
inline double &RF(double &a, double &b, double &c, double &d, double &e,
                  double &f) {
  RF(a), RF(b), RF(c), RF(d), RF(e), RF(f);
  return a;
}
inline double &RF(double &a, double &b, double &c, double &d, double &e,
                  double &f, double &g) {
  RF(a), RF(b), RF(c), RF(d), RF(e), RF(f), RF(g);
  return a;
}
inline void RS(char *s1, char *s2) { RS(s1), RS(s2); }
inline void RS(char *s1, char *s2, char *s3) { RS(s1), RS(s2), RS(s3); }
template <class T>
inline void RST(T &A) {
  memset(A, 0, sizeof(A));
}
template <class T>
inline void FLC(T &A, int x) {
  memset(A, x, sizeof(A));
}
template <class T>
inline T &SRT(T &A) {
  sort(A.begin(), A.end());
  return A;
}
template <class T, class C>
inline T &SRT(T &A, C B) {
  sort(A.begin(), A.end(), B);
  return A;
}
template <class T>
inline T &UNQ(T &A) {
  A.resize(unique(SRT(A).begin(), SRT(A).end()) - A.begin());
  return A;
}
const int dx4[] = {-1, 0, 1, 0};
const int dy4[] = {0, 1, 0, -1};
const int dx8[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy8[] = {0, 1, 0, -1, -1, 1, -1, 1};
const int dxhorse[] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dyhorse[] = {1, -1, 2, -2, 2, -2, 1, -1};
const int MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long INFF = 1LL << 60;
const double EPS = 1e-9;
const double OO = 1e15;
const double PI = acos(-1.0);
template <class T>
inline void checkMin(T &a, const T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkMax(T &a, const T b) {
  if (a < b) a = b;
}
template <class T>
inline void checkMin(T &a, T &b, const T x) {
  checkMin(a, x), checkMin(b, x);
}
template <class T>
inline void checkMax(T &a, T &b, const T x) {
  checkMax(a, x), checkMax(b, x);
}
template <class T, class C>
inline void checkMin(T &a, const T b, C c) {
  if (c(b, a)) a = b;
}
template <class T, class C>
inline void checkMax(T &a, const T b, C c) {
  if (c(a, b)) a = b;
}
template <class T>
inline T min(T a, T b, T c) {
  return min(min(a, b), c);
}
template <class T>
inline T max(T a, T b, T c) {
  return max(max(a, b), c);
}
template <class T>
inline T min(T a, T b, T c, T d) {
  return min(min(a, b), min(c, d));
}
template <class T>
inline T max(T a, T b, T c, T d) {
  return max(max(a, b), max(c, d));
}
template <class T>
inline T sqr(T a) {
  return a * a;
}
template <class T>
inline T cub(T a) {
  return a * a * a;
}
inline int ceil(int x, int y) { return (x - 1) / y + 1; }
inline int sgn(double x) { return x < -EPS ? -1 : x > EPS; }
inline int sgn(double x, double y) { return sgn(x - y); }
namespace BO {
inline bool _1(int x, int i) { return bool(x & 1 << i); }
inline bool _1(long long x, int i) { return bool(x & 1LL << i); }
inline long long _1(int i) { return 1LL << i; }
template <class T>
inline bool odd(T x) {
  return x & 1;
}
template <class T>
inline bool even(T x) {
  return !odd(x);
}
template <class T>
inline T low_bit(T x) {
  return x & -x;
}
inline int count_bits(int x) { return __builtin_popcount(x); }
inline int count_bits(long long x) { return __builtin_popcountll(x); }
}  // namespace BO
using namespace BO;
namespace NT {
inline void INC(int &a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
inline int sum(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
inline void DEC(int &a, int b) {
  a -= b;
  if (a < 0) a += MOD;
}
inline int dff(int a, int b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
inline void MUL(int &a, int b) { a = (long long)a * b % MOD; }
inline int pdt(int a, int b) { return (long long)a * b % MOD; }
inline int sum(int a, int b, int c) { return sum(sum(a, b), c); }
inline int sum(int a, int b, int c, int d) { return sum(sum(a, b), sum(c, d)); }
inline int pdt(int a, int b, int c) { return pdt(pdt(a, b), c); }
inline int pdt(int a, int b, int c, int d) { return pdt(pdt(pdt(a, b), c), d); }
inline int pow(int a, int b) {
  int c(1);
  while (b) {
    if (b & 1) MUL(c, a);
    MUL(a, a), b >>= 1;
  }
  return c;
}
inline int pow(int a, long long b) {
  int c(1);
  while (b) {
    if (b & 1) MUL(c, a);
    MUL(a, a), b >>= 1;
  }
  return c;
}
template <class T>
inline T pow(T a, long long b) {
  T c(1);
  while (b) {
    if (b & 1) c *= a;
    a *= a, b >>= 1;
  }
  return c;
}
inline int _I(int b) {
  int a = MOD, x1 = 0, x2 = 1, q;
  while (true) {
    q = a / b, a %= b;
    if (!a) return (x2 + MOD) % MOD;
    DEC(x1, pdt(q, x2));
    q = b / a, b %= a;
    if (!b) return (x1 + MOD) % MOD;
    DEC(x2, pdt(q, x1));
  }
}
inline void DIV(int &a, int b) { MUL(a, _I(b)); }
inline int qtt(int a, int b) { return pdt(a, _I(b)); }
inline int phi(int n) {
  int res = n;
  for (int i = 2; sqr(i) <= n; ++i)
    if (!(n % i)) {
      DEC(res, qtt(res, i));
      do {
        n /= i;
      } while (!(n % i));
    }
  if (n != 1) DEC(res, qtt(res, n));
  return res;
}
}  // namespace NT
using namespace NT;
namespace Math {
long long GCD(long long a, long long b) { return b ? GCD(b, a % b) : a; }
long long extendGCD(long long a, long long b, long long &x, long long &y) {
  if (!b) return x = 1, y = 0, a;
  long long res = extendGCD(b, a % b, x, y), tmp = x;
  x = y, y = tmp - (a / b) * y;
  return res;
}
long long power(long long x, long long k) {
  long long res = 1;
  while (k) {
    if (k & 1) res *= x;
    x *= x, k >>= 1;
  }
  return res;
}
long long powerMod(long long x, long long k, long long m) {
  long long res = 1;
  while (x %= m, k) {
    if (k & 1) res *= x, res %= m;
    x *= x, k >>= 1;
  }
  return res;
}
long long inverse(long long a, long long p, long long t = 1) {
  long long pt = power(p, t);
  long long x, y;
  y = extendGCD(a, pt, x, y);
  return x < 0 ? x += pt : x;
}
long long linearCongruence(long long a, long long b, long long p, long long q) {
  long long x, y;
  y = extendGCD(p, q, x, y);
  while (b < a) b += q / y;
  x *= b - a, x = p * x + a, x %= p * q;
  if (x < 0) x += p * q;
  return x;
}
const int PRIMERANGE = 1000000;
int prime[PRIMERANGE + 1];
int getPrime() {
  memset(prime, 0, sizeof(int) * (PRIMERANGE + 1));
  for (int i = 2; i <= PRIMERANGE; i++) {
    if (!prime[i]) prime[++prime[0]] = i;
    for (int j = 1; j <= prime[0] && prime[j] <= PRIMERANGE / i; j++) {
      prime[prime[j] * i] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  return prime[0];
}
int factor[100][3], facCnt;
int getFactors(int x) {
  facCnt = 0;
  int tmp = x;
  for (int i = 1; prime[i] <= tmp / prime[i]; i++) {
    factor[facCnt][1] = 1, factor[facCnt][2] = 0;
    if (tmp % prime[i] == 0) factor[facCnt][0] = prime[i];
    while (tmp % prime[i] == 0)
      factor[facCnt][2]++, factor[facCnt][1] *= prime[i], tmp /= prime[i];
    if (factor[facCnt][1] > 1) facCnt++;
  }
  if (tmp != 1)
    factor[facCnt][0] = tmp, factor[facCnt][1] = tmp, factor[facCnt++][2] = 1;
  return facCnt;
}
long long combinationModP(long long n, long long k, long long p) {
  if (k > n) return 0;
  if (n - k < k) k = n - k;
  long long a = 1, b = 1, x, y;
  int pcnt = 0;
  for (int i = 1; i <= k; i++) {
    x = n - i + 1, y = i;
    while (x % p == 0) x /= p, pcnt++;
    while (y % p == 0) y /= p, pcnt--;
    x %= p, y %= p, a *= x, b *= y;
    b %= p, a %= p;
  }
  if (pcnt) return 0;
  extendGCD(b, p, x, y);
  if (x < 0) x += p;
  a *= x, a %= p;
  return a;
}
};  // namespace Math
namespace Geo {
const double eps = 1e-8;
int dblcmp(double d) { return d < -eps ? -1 : d > eps; }
int sgn(double a) { return a < -eps ? -1 : a > eps; }
inline double sqr(double x) { return x * x; }
};  // namespace Geo
template <class T>
inline T &RD(T &x) {
  char c;
  for (c = getchar(); c < '-'; c = getchar())
    ;
  if (c == '-') {
    x = '0' - getchar();
    for (c = getchar(); '0' <= c && c <= '9'; c = getchar())
      x = x * 10 + '0' - c;
  } else {
    x = c - '0';
    for (c = getchar(); '0' <= c && c <= '9'; c = getchar())
      x = x * 10 + c - '0';
  }
  return x;
}
inline double &RF(double &x) {
  scanf("%lf", &x);
  return x;
}
inline char *RS(char *s) {
  scanf("%s", s);
  return s;
}
int Case;
template <class T>
inline void OT(const T &x) {
  cout << x << endl;
}
int n;
void solve() {
  cin >> n;
  vector<int> p, q;
  p.clear();
  q.clear();
  cout << 1 << ' ' << n - 1 << ' ' << 1;
  for (int i = 2; i <= n; ++i) {
    p.push_back(i);
    cout << ' ' << i;
  }
  cout << endl;
  cout.flush();
  int d, dd;
  cin >> d;
  while (p.size() > 1) {
    q.clear();
    while (p.size() - 1 >= q.size() + 1) {
      q.push_back(p.back());
      p.pop_back();
    }
    cout << 1 << ' ' << int(q.size()) << ' ' << 1;
    for (int i = 0; i < int(q.size()); ++i) cout << ' ' << q[i];
    cout << endl;
    cout.flush();
    cin >> dd;
    if (dd == d) {
      p = q;
    }
  }
  int x = p.back();
  cout << 1 << ' ' << n - 1 << ' ' << x;
  for (int i = 1; i <= n; ++i)
    if (i != x) {
      cout << ' ' << i;
    }
  cout << endl;
  cout.flush();
  cin >> d;
  cout << -1 << ' ' << d << endl;
  cout.flush();
}
int main() {
  cin >> Case;
  while (Case--) solve();
}