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 <iostream>
#include <vector>
#include <utility>

int main() {
		
	// recieve |V| and |E|
	int V, E; std::cin >> V >> E;
	std::vector<std::vector<std::pair<int, int>>> G(V);
	
	for(int i = 0; i < E; i++) {
		// recieve edges
		int u, v, d; std::cin >> u >> v >> d;
		u--; v--;
		G[u].push_back({v, d});
		G[v].push_back({u, d});
	}

	// calc cost
	std::vector<std::vector<int>> D(V,std::vector<int>(V));
	for(int i = 0; i < V; i++){
		for(int j = 0; j < V; j++){
			D[i][j] = 999999999;
		}
		std::vector<bool> flag(V);
		D[i][i] = 0;
		for(int j = 0; j < V; j++){
			int minnode = -1;
			int mincost = 999999999;
			for(int k = 0; k < V; k++){
				if(flag[k]==false && D[i][k] < mincost){
					minnode = k;
					mincost = D[i][k];
				}
			}
			flag[minnode] = true;
			for(int k = 0; k < G[minnode].size(); k++){
				if(D[i][minnode]+G[minnode][k].second < D[i][G[minnode][k].first]){
					D[i][G[minnode][k].first] = D[i][minnode]+G[minnode][k].second;
				}
			}
		}
	}
	
	int Tmax; std::cin >> Tmax;
	long long score = 0;
	std::vector<int> orderFlag;
	std::vector<std::pair<int, int>> ORDER;

	int nowNode = 0;
	int targetNode = 0;
	int remainDist = 0;

	// recieve info
	for(int i = 0; i < Tmax; i++) {
		int Nnew; std::cin >> Nnew;
		for(int j = 0; j < Nnew; j++) {
			int new_id, dst; std::cin >> new_id >> dst;
			dst--;
			orderFlag.push_back(0);
			ORDER.push_back({i, dst});
		}
		// insert your code here to get more meaningful output
		int ans = -1;
		if(remainDist == 0){
			nowNode = targetNode;
			int mintarget = -1;
			int mincost = 999999999;
			for(int j = 0; j < ORDER.size(); j++){
				if(orderFlag[j] == 1 && nowNode == ORDER[j].second){
					score += Tmax * Tmax - (i - ORDER[j].first) * (i - ORDER[j].first);
					orderFlag[j] = 2;
				}
				if(nowNode == 0 && orderFlag[j] == 0){
					orderFlag[j] = 1;
				}
				if(orderFlag[j] == 0){
					if(D[nowNode][0]+D[0][ORDER[j].second] <= mincost){
						mintarget = 0;
						mincost = D[nowNode][ORDER[j].second];
					}
				}
				if(orderFlag[j] == 1){
					if(D[nowNode][ORDER[j].second] < mincost){
						mintarget = ORDER[j].second;
						mincost = D[nowNode][ORDER[j].second];
					}
				}
			}
			if(mintarget == -1 && nowNode > 0){
				mintarget = 0;
			}
			int minNextCost = 999999999;
			if(mintarget >= 0){
				for(int j = 0; j < G[nowNode].size(); j++){
					if(D[G[nowNode][j].first][mintarget] < minNextCost){
						ans = G[nowNode][j].first + 1;
						targetNode = ans - 1;
						remainDist = G[nowNode][j].second - 1;
						minNextCost = D[G[nowNode][j].first][mintarget];
					}
				}
			}
		} else {
			ans = targetNode + 1;
			remainDist--;
        }
		std::cout << ans << std::endl;
		//std::cerr << nowNode << " " << targetNode << " " << remainDist << std::endl;
	}
	std::cerr << score << std::endl;
	return 0;
}