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;
#ifdef _DEBUG
#define _GLIBCXX_DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

#define int long long
#define ll long long
#define DBG 1
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define rrep(i, a, b) for (int i = (b)-1; i >= (a); i--)
#define loop(n) rep(loop, (0), (n))
#define all(c) begin(c), end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
using pii = pair<int, int>;
// template<class T> ostream &operator<<(ostream &os,T &t){dump(t);return os;}
template <typename T, typename S>istream &operator>>(istream &is, pair<T, S> &p) { is >> p.first >> p.second; return is; }
//template <typename T, typename S>ostream &operator<<(ostream &os, pair<T, S> &p) {os << p.first << " " << p.second;return os;}

template <typename T> void printvv(const vector<vector<T>> &v) {
	cerr << endl;
	rep(i, 0, v.size()) rep(j, 0, v[i].size()) {
		if (typeid(v[i][j]).name() == typeid(INF).name() and v[i][j] == INF) {
			cerr << "INF";
		}
		else
			cerr << v[i][j];
		cerr << (j == v[i].size() - 1 ? '\n' : ' ');
	}
	cerr << endl;
}

#ifndef _DEBUG
#define printvv(...)
#endif
void YES(bool f) { cout << (f ? "YES" : "NO") << endl; }
void Yes(bool f) { cout << (f ? "Yes" : "No") << endl; }
template <class T, class U>bool chmin(T& a, U b) { if (a > b) { a = b; return true; }return false; }
template <class T, class U>bool chmax(T& a, U b) { if (a < b) { a = b; return true; }return false; }
using Weight = int;
using Flow = int;
struct Edge {
	int s, d; Weight w; Flow c; int id;
	Edge() {};
	Edge(int s, int d, Weight w = 1) : s(s), d(d), w(w), c(w) {};
};
bool operator<(const Edge &e1, const Edge &e2) { return e1.w < e2.w; }
bool operator>(const Edge &e1, const Edge &e2) { return e2 < e1; }
inline ostream &operator<<(ostream &os, const Edge &e) { return (os << '(' << e.s << ", " << e.d << ", " << e.w << ')'); }

using Edges = vector<Edge>;
using Graph = vector<Edges>;
using Array = vector<Weight>;
using Matrix = vector<Array>;

void addArc(Graph &g, int s, int d, Weight w = 1) {
	g[s].emplace_back(s, d, w);
}
void addEdge(Graph &g, int a, int b, Weight w = 1) {
	addArc(g, a, b, w);
	addArc(g, b, a, w);
}
//単一始点最短経路(負閉路なし)
//Dijkstra O((|E|+|V|)log|V|)
//dist: 始点から各頂点までの最短距離
//戻り値: 最短経路木の親頂点(根は-1)
vector<int> dijkstra(const Graph &g, int s, Array &dist) {
	int n = g.size();
	assert(s < n);
	enum { WHITE, GRAY, BLACK };
	vector<int> color(n, WHITE); color[s] = GRAY;
	vector<int> prev(n, -1);
	dist.assign(n, INF); dist[s] = 0;
	using State = tuple<Weight, int, int>; //始点からの最短距離 子 親
	priority_queue<State, vector<State>, greater<State>> pq; pq.emplace(0, s, -1);
	while (pq.size()) {
		Weight d; int v, u; tie(d, v, u) = pq.top(); pq.pop();
		if (dist[v] < d)continue;
		color[v] = BLACK; prev[v] = u;
		for (auto &e : g[v]) {
			if (color[e.d] == BLACK)continue;
			if (dist[e.d] > dist[v] + e.w) {
				dist[e.d] = dist[v] + e.w;
				pq.emplace(dist[e.d], e.d, v);
				color[e.d] = GRAY;
			}
		}
	}
	return prev;
}

struct Info {
	int id;
	int dst;
	int time;
	Info(int id, int dst, int time) :id(id), dst(dst), time(time) {}
};
int V, E;
Edges edges;
int Tmax;
vector<vector<Info>>infos;
void Input() {
	cin >> V >> E;
	loop(E) {
		int a, b, c; cin >> a >> b >> c;
		a--, b--;
		edges.eb(a, b, c);
	}
	cin >> Tmax;
	infos.resize(Tmax);
	rep(i, 0, Tmax) {
		int n; cin >> n;
		rep(j, 0, n) {
			int id, d; cin >> id >> d;
			id--; d--;
			infos[i].eb(id, d, i);
		}
	}
}

void solve() {
	Graph g(V);
	for (auto e : edges) {
		addEdge(g, e.s, e.d, e.w);
	}
	Matrix dist(V);
	vector<vector<int>>prev(V);
	rep(i, 0, V) {
		prev[i] = dijkstra(g, i, dist[i]);
	}

	vector<string>cmds;
	int root = 0;
	int info_idx = 0;
	queue<Info>que;
	auto move = [&](int s, int d) {
		if (s == d)return;
		int cur = s;
		vector<int>path;
		int v = d;
		while (v != cur) {
			path.eb(v);
			v = prev[cur][v];
		}
		reverse(all(path));
		for (auto v : path) {
			int i;
			for (i = 0; i < dist[cur][v] and cmds.size() < Tmax; i++) {
				cmds.eb(to_string(v + 1));
			}
			if (i == dist[cur][v]) {
				cur = v;
			}
		}
	};
	while (cmds.size() < Tmax) {
		int t = cmds.size();
		while (info_idx <= t) {
			for (auto info : infos[info_idx]) {
				que.emplace(info);
			}
			info_idx++;
		}
		if (que.empty()) {
			cmds.eb("-1");
			continue;
		}
		int cur = root;
		while (que.size() and cmds.size() < Tmax) {
			auto info = que.front(); que.pop();
			if (dist[cur][info.dst] > Tmax - cmds.size())continue;
			move(cur, info.dst);
			cur = info.dst;
		}
		if (dist[cur][root] > Tmax - cmds.size()) {
			while (cmds.size() < Tmax) {
				cmds.eb("-1");
			}
			break;
		}
		move(cur, root);
		cur = root;
	}

	assert(cmds.size() == Tmax);
	for (auto s : cmds) {
		cout << s << endl;
	}
}

signed main(signed argc, char *argv[]) {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout << fixed << setprecision(12);

	Input();
	solve();

	return 0;
}