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;

static const int INFTY = 1e+9;
static const int WHITE = 0;
static const int GRAY = 1;
static const int BLACK = 2;
static const int VERTEX_MAX = 801;
static const int MAX_ORDER = 10001;

int cost_matrix[VERTEX_MAX][VERTEX_MAX];
int V, E, u, v, cost, Tmax, Nnew, new_id, dst;
int d[VERTEX_MAX], via[VERTEX_MAX];
int output_count = 0;

struct shop
{
	int shop_position = 1;
	deque<tuple<int, int, int>> order;
} Shop;

void dijkstra(int start, int end)
{
	int minv;
	int color[VERTEX_MAX];
	
	for (int i = 0; i < V+1; ++i)
	{
		d[i] = INFTY;
		color[i] = WHITE;
		via[i] = -1;
	}
	
	d[start] = 0;
	color[start] = GRAY;
	
	while(1)
	{
		minv = INFTY;
		int u = -1;
		
		for (int i = 0; i < V+1; ++i)
		{
			if (color[i] != BLACK && d[i] < minv)
			{
				u = i;
				minv = d[i];
			}
		}
		
		if (u == end)
		{
			break;
		}
		
		color[u] = BLACK;
		
		for (int v = 0; v < V+1; ++v)
		{
			if (color[v] != BLACK && cost_matrix[u][v] != INFTY)
			{
				if (d[v] > d[u] + cost_matrix[u][v])
				{
					d[v] = d[u] + cost_matrix[u][v];
					color[v] = GRAY;
					via[v] = u;
				}
			}
		}
	}
}

void delivery(int start, int end)
{
	dijkstra(start, end);
	
	vector<int> start_to_end;
	int node = end;
	start_to_end.push_back(node);
	
	while(1)
	{
		if (start == end)
		{
			break;
		}
		
		node = via[node];
		start_to_end.push_back(node);
		
		if (node == start)
		{
			break;
		}
	}
	
	reverse(start_to_end.begin(), start_to_end.end());
	
	int next_cost[VERTEX_MAX];
	for (int i = 1; i < start_to_end.size(); ++i)
	{
		next_cost[i] = d[start_to_end[i]] - d[start_to_end[i-1]];
		
		for (int j = 0; j < next_cost[i]; ++j)
		{
			if (output_count >= Tmax)
			{
				break;
			}
			cout << start_to_end[i] << endl;
			output_count++;
		}
	}
}

int main() 
{
	cin >> V >> E;
	
	for (int i = 0; i < E; ++i)
	{
		for (int j = 0; j < E; ++j)
		{
			cost_matrix[i][j] = INFTY;
		}
	}
	
	for(int i = 0; i < E; i++)
	{
		cin >> u >> v >> cost;
		
		cost_matrix[u][v] = cost;
		cost_matrix[v][u] = cost;
	}
	
	cin >> Tmax;
	
	for(int i = 0; i < Tmax; i++)
	{
		cin >> Nnew;
		
		for(int j = 0; j < Nnew; j++)
		{
			cin >> new_id >> dst;
			
			Shop.order.emplace_back(i, new_id, dst);
		}
	}
	
	for (int t = 0; t < get<0>(Shop.order[0]); ++t)
	{
		cout << "-1" << endl;
		output_count++;
	}
	
	int order_update = 0;
	int start = Shop.shop_position;
	int end;
	
	for (int t = get<0>(Shop.order[0]); t < Tmax; ++t)
	{
		end = get<2>(Shop.order[order_update]);
		
		delivery(start, end);
		
		delivery(end, start);
		
		if (output_count >= Tmax)
		{
			break;
		}
		
		order_update++;
	}
	
	return 0;
}