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
#include <bits/stdc++.h>
#define pb push_back
#define sz(a) (int)(a.size())
#define x first
#define y second
using namespace std;
const int MAXV = 405;
const int MAXT = 10005;
vector<pair<int, int> > g[MAXV];
int id[MAXT] , dst[MAXT] , Tmax , n , m;
void inputG()
{
	cin >> n >> m;
	for(int i = 0; i < m; i++) {
		int u, v, d; cin >> u >> v >> d;
		u--; v--;
		g[u].push_back({v, d});
		g[v].push_back({u, d});
	}
}
void inputQueries()
{
	
	cin >> Tmax;
	memset(id , -1 , sizeof id);
	for(int i = 0; i < Tmax; i++) {
		int Nnew; cin >> Nnew;
		for(int j = 0; j < Nnew; j++) {
			cin >> id[i] >> dst[i];
		}
	}
}
vector<pair<int , int> > spanTree[MAXV];
int p[MAXV];
int find(int u)
{
	return p[u] = p[u] == u ? u : find(p[u]);
}
void unite(int u , int v , int c)
{
	int U = find(u) , V = find(v);
	if(U != V)
	{
		p[V] = U;
		spanTree[u].pb({v , c});
		spanTree[v].pb({u , c});
	}
}

void buildSpan()
{
	for(int u = 0; u < n; u++)p[u] = u;
	vector<pair<int , pair<int , int> > > edges;
	for(int u = 0; u < n; u++)
		for(auto v : g[u])
			if(u < v.x)
				edges.pb({v.y , {u , v.x}});
	sort(edges.begin() , edges.end());
	for(auto e : edges)
		unite(e.y.x , e.y.y , e.x);
}

vector<int> route;
void dfs(int u , int p = -1)
{
	for(auto v : spanTree[u])
		if(v.x != p)
		{
			for(int i = 0; i < v.y; i++)
				route.pb(v.x);
			dfs(v.x , u);
			for(int i = 0; i < v.y; i++)
				route.pb(u);
		}
			
}
void findRoute()
{
	dfs(0);
}
int main() {
	inputG();
	inputQueries();
	buildSpan();
	findRoute();
	int cntStop = 400;
	for(int i = 0; i < Tmax; i++)
	{
		if(i < cntStop)
			cout << -1 << endl;
		else
			cout << route[(i - cntStop) % sz(route)] + 1 << endl;
	}
	return 0;
}
