Problem: Requirements

* All inputs are of non-negative integer value.
* T_{\text{max}} = 10000
* 200 \leq |V| \leq 400
* 1.5 |V| \leq |E| \leq 2 |V|
* 1 \leq u_{i}, v_{i} \leq |V| (1 \leq i \leq |E|)
* 1 \leq d_{u_i, v_i} \leq \lceil 4\sqrt{2|V|} \rceil (1 \leq i \leq |E|)
* The given graph has no self-loops / multiple edges and is guaranteed to be connected.
* 0 \leq N_{\text{new}} \leq 1
* 1 \leq \mathrm{new\\_id}_{i} \leq T_{\text{last}}+1 (1 \leq i \leq N_{\text{new}}). Note: If all orders are generated by the order generation rule as explained above, the total number of orders is at most T_{\text{last}}+1. Therefore, the possible range of \mathrm{new\\_id}_{i} should be from 1 to T_{\text{last}}+1.
* 2 \leq \mathrm{dst}_i \leq |V| (1 \leq i \leq N_{\text{new}})
* The order IDs \mathrm{new\\_{id}}_i are unique.

Input Format

Input is provided in the following form:


|V| |E|
u_{1} v_{1} d_{u_{1}, v_{1}}
u_{2} v_{2} d_{u_{2}, v_{2}}
:
u_{|E|} v_{|E|} d_{u_{|E|}, v_{|E|}}
T_{\text{max}}
\mathrm{info}_{0}
\mathrm{info}_{1}
:
\mathrm{info}_{T_{\text{max}}-1}


* In the first line |V| denotes the number of vertices, while |E| denotes the number of edges.
* The following |E| lines denote the edges of the graph. In particular, in the ith line u_{i} and v_{i} denote the edge connecting u_{i} and v_{i} and d_{u_{i}, v_{i}} the corresponding distance.
* The following line denotes the number of steps T_{\text{max}}.



In the following line, \mathrm{info}_t is information about the order from the customer that occurs at time t. \mathrm{info}_t is given in the form:


N_{\text{new}}
\mathrm{new\_id}_1 \mathrm{dst}_1
\mathrm{new\_id}_2 \mathrm{dst}_2
\vdots
\mathrm{new\_id}_{N_{\text{new}}} \mathrm{dst}_{N_{\text{new}}}


* N_{\text{new}} represents the number of new orders which appear at time t.
* The next N_{\text{new}} lines give the newly generated order information. The i-th order information indicates that the order ID \mathrm{new\\_{id}}_i of the new order, while \mathrm{dst}_i denotes the vertex to which the customer wishes the order to be delivered.
* Note: If N_{\text{new}}=0, there are no new lines.

Output Format

The Output expects T_{\text{max}} integers in the format specified below.


\mathrm{command}_{0}
\mathrm{command}_{1}
:
\mathrm{command}_{T_{\text{max}}-1}


In particular, \mathrm{command}_{i} shall specify the movement of the delivery car by using one of the following two options:

1) `stay`, if the car shall not move:


-1


2) `move w`, if the car shall be moved one step towards the neighboring vertex w


w


Note that in case of 2) w has to satisfy the following conditions:

* w \in V
* If the car is at vertex u: \left\\{ u, w \right\\} \in E .
* If the car is on the edge \left\\{ u, v \right\\}, w must either satisfy u = w or v = w.



* * *

Examples

Input




Output




Input

5 7
1 2 5
5 3 4
2 4 8
1 5 1
2 3 3
4 5 3
4 3 9
4
1
1 2
1
2 5
1
3 4
0


Output

2
-1
1
5

Solution (in CPP):

Solution:
#include <algorithm>
#include <chrono>
#include <iostream>
#include <memory>
#include <random>
#include <vector>
using namespace std;

struct Order {
  int t;
  int id;
  int dest;
};

struct State {
  State() = default;
  explicit State(State* other)
    : t(other->t),
      position(-1),
      orders(other->orders),
      pick_up(other->pick_up),
      score(other->score),
      prev(other) {}

  int t = 0;
  int position = 0;
  vector<int> orders;
  int pick_up = 0;

  int score = 0;
  double estimation = 0;
  shared_ptr<State> prev;
};

int main() {
  int V, E;
  cin >> V >> E;
  vector<vector<int>> distance(V, vector<int>(V, -1));
  vector<vector<int>> neighbors(V);
  for (int i = 0; i < V; ++i)
    distance[i][i] = 0;
  for (int i = 0; i < E; ++i) {
    int u, v, d;
    cin >> u >> v >> d;
    --u;
    --v;
    distance[u][v] = d;
    distance[v][u] = d;
    neighbors[u].push_back(v);
    neighbors[v].push_back(u);
  }

  int T_max;
  cin >> T_max;
  vector<Order> orders;
  for (int t = 0; t < T_max; ++t) {
    int N;
    cin >> N;
    for (int i = 0; i < N; ++i) {
      int id, dest;
      cin >> id >> dest;
      orders.emplace_back(Order {t, id, dest - 1});
    }
  }

  mt19937 prg;
  int pos = 0;
  for (int t = 0; t < T_max;) {
    int next = neighbors[pos][prg() % neighbors[pos].size()];
    int d = distance[pos][next];
    for (int i = 0; i < d && t + i < T_max; ++i)
      cout << next + 1 << "\n";
    pos = next;
    t += d;
  }

  return 0;
}