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;

class state{
public:
	int vi, eij, d;
	state(int _vi, int _eij, int _d) : vi(_vi), eij(_eij), d(_d) {};
};
class edge{
public:
	int cost;
	size_t from, to;
	edge(int c, size_t f, size_t t){
		cost = c;
		from = f;
		to = t;
	}
};
using edges = vector<edge>;
using graph = vector<edges>;
void readgraph(graph &g, size_t E, bool D, bool C, bool Z){ // E:num_of_edges, D:directed, C:costed, Z:already 0-indexed
	for(size_t i=0; i<E; ++i){
		edge e(1, 0, 0);
		if(!C) cin >> e.from >> e.to;
		else cin >> e.from >> e.to >> e.cost;
		if(!Z){
			--e.from;
			--e.to;
		}
		g[e.from].push_back(e);
		if(!D){
			swap(e.from, e.to);
			g[e.from].push_back(e);
		}
	}
}
bool operator<(const edge &e, const edge &f){
	return e.cost != f.cost ? e.cost > f.cost : e.from != f.from ? e.from < f.from : e.to < f.to;
}
void dijkstra(const graph &g, size_t s, vector<int> &dist){
	size_t V = g.size();
	dist = vector<int>(V, 1e8);
	dist[s] = 0;
	priority_queue<edge> q;
	q.push(edge(0, s, s));
	while(!q.empty()){
		edge e = q.top(); q.pop();
		if(dist[e.to] < e.cost) continue;
		for(const edge f:g[e.to]){
			if(dist[f.to] > dist[e.to] + f.cost){
				dist[f.to] = dist[e.to] + f.cost;
				q.push(edge(dist[f.to], s, f.to));
			}
		}
	}
}

int v, e, tmax;
graph g;
vector<vector<int>> d, h;
using P = pair<int, int>;
vector<P> info;

void input(){
	cin >> v >> e;
	g.resize(v);
	readgraph(g, e, false, true, false);
	cin >> tmax;
	info.resize(tmax, P(-1, -1));
	for(int i=0; i<tmax; ++i){
		int nnew; cin >> nnew;
		if(nnew == 1){
			int id, dst; cin >> id >> dst; --dst;
			info[i] = P(id, dst);
		}
	}
}
void calc_h(){
	d.resize(v);
	for(int i=0; i<v; ++i) dijkstra(g, i, d[i]);
	h.resize(v, vector<int>(v, -1));
	for(int i=0; i<v; ++i)
		for(int j=0; j<v; ++j)
			for(size_t k=0; k<g[i].size(); ++k)
				if(h[i][j] == -1 || d[g[i][h[i][j]].to][j] > d[g[i][k].to][j])
					h[i][j] = k;
}
void solve(){
	state s(0, 0, 0);
	priority_queue<P> pq;
	for(int i=0; i<min(tmax, 10); ++i){
		if(info[i] != P(-1, -1)) pq.push(P(tmax-i, info[i].second));
		cout << -1 << "\n";
	}
	pq.push(P(tmax-9, 0));
	for(int i=10; i<tmax; ++i){
		if(i % 100 == 0) pq.push(P(tmax-i+100, 0));
		if(info[i] != P(-1, -1)) pq.push(P(tmax-i, info[i].second));
		if(pq.empty()){
			cout << -1 << "\n";
			continue;
		}
		P p = pq.top();
		if(s.vi == p.second){
			cout << -1 << "\n";
			if(!pq.empty()) pq.pop();
			continue;
		}
		if(s.d == 0) s.eij = h[s.vi][p.second];
		cout << g[s.vi][s.eij].to+1 << "\n";
		++s.d;
		if(s.d == g[s.vi][s.eij].cost){
			s.vi = g[s.vi][s.eij].to;
			s.eij = 0;
			s.d = 0;
		}
		if(s.vi == p.second && !pq.empty()) pq.pop();
	}
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	input();
	calc_h();
	solve();
}