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 <bits/stdc++.h>
using namespace std;

constexpr int INF = 1e9;
constexpr int STORE = 0;

struct Order {
  int t, n, id, dst;
};

struct Position {
  int u, v, du, dv;
};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  // input
  int V, E, T;
  cin >> V >> E;
  vector<vector<int>> adjacency_matrix(V, vector<int>(V, INF));
  for (int i = 0; i < E; i++) {
    int u, v, d;
    cin >> u >> v >> d;
    u--, v--;
    adjacency_matrix[u][v] = adjacency_matrix[v][u] = d;
  }
  cin >> T;
  vector<Order> orders(T);
  for (int i = 0; i < T; i++) {
    orders[i].t = i;
    cin >> orders[i].n;
    if (orders[i].n) {
      cin >> orders[i].id >> orders[i].dst;
      orders[i].dst--;
    }
  }

  // solve
  // Warshall-Floyd algorithm
  vector<vector<int>> distance_matrix = adjacency_matrix;
  for (int i = 0; i < V; i++) distance_matrix[i][i] = 0;
  vector<vector<int>> next_node_matrix(V, vector<int>(V));
  for (int i = 0; i < V; i++) {
    for (int j = 0; j < V; j++) {
      next_node_matrix[i][j] = j;
    }
  }
  for (int k = 0; k < V; k++) {
    for (int i = 0; i < V; i++) {
      for (int j = 0; j < V; j++) {
        if (distance_matrix[i][k] + distance_matrix[k][j] < distance_matrix[i][j]) {
          distance_matrix[i][j] = distance_matrix[i][k] + distance_matrix[k][j];
          next_node_matrix[i][j] = next_node_matrix[i][k];
        }
      }
    }
  }

  vector<int> commands(T);
  Position position{STORE, STORE, 0, 0};
  queue<int> destinations;
  queue<Order> produced_orders;
  for (int t = 0; t < T; t++) {
    if (orders[t].n) {
      produced_orders.push(orders[t]);
    }
    if (position.du == 0 && position.u == STORE || position.dv == 0 && position.v == STORE) {
      while (!produced_orders.empty()) {
        destinations.push(produced_orders.front().dst);
        produced_orders.pop();
      }
      destinations.push(STORE);
    }

    if (position.du == 0) {
      while (!destinations.empty() && destinations.front() == position.u) destinations.pop();
    } else if (position.dv == 0) {
      while (!destinations.empty() && destinations.front() == position.v) destinations.pop();
    }
    if (destinations.empty()) {
      commands[t] = -1;
    } else if (position.du == 0 || position.dv == 0) {
      int i = position.du == 0 ? position.u : position.v;
      int j = next_node_matrix[i][destinations.front()];
      commands[t] = j;
      position = {i, j, 1, adjacency_matrix[i][j] - 1};
    } else {
      if (position.du + distance_matrix[position.u][destinations.front()] <
          position.dv + distance_matrix[position.v][destinations.front()]) {
        commands[t] = position.u;
        position.du--;
        position.dv++;
      } else {
        commands[t] = position.v;
        position.du++;
        position.dv--;
      }
    }
  }

  // output
  for (auto e : commands) cout << e + (e != -1) << '\n';
  cout << flush;
}