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<string>
#include<queue>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<math.h>

using namespace std;

#define INF 1000000007
#define LINF (1LL << 62)

typedef long long i64;
typedef pair<i64,i64> P;


i64 T, n, m, to[10101];
vector<int> v[404];
vector<P> G[404];
vector<vector<int>> dist(404, vector<int>(404, INF));
vector<int> route[404][404];
i64 now = 1, next_t = 0;
i64 have[404] = {0};

vector<i64> result;
int finished[404];
i64 o[100][10001];
int Times = 0;

i64 para[2] = {180, 16};

void floyd(){
	for(int i = 1; i <= n; i++){
		for(int j = 1; j <= n; j++){
			for(int k = 1; k <= n; k++){
				if(dist[j][k] > dist[j][i] + dist[i][k]){
					//route[j][k] = route[j][i]+route[i][k];
					route[j][k] = route[j][i];
					route[j][k].insert(route[j][k].end(),route[i][k].begin(),route[i][k].end());
					dist[j][k] = dist[j][i] + dist[i][k];
				}
			}
		}
	}
}

P choice_next(int start, i64 t){
	i64 ma = -1, k = -1;
	for(int i = 2; i <= n; i++){
		if(!have[i]) continue;
		if(dist[start][i] >= T-t) continue;
		i64 cnt = 0;
		for(int j = 0; j < route[start][i].size(); j++){
			cnt += have[route[start][i][j]];
		}
		if(ma < cnt * para[0] / (dist[start][i])){
			ma = cnt * para[0] / (dist[start][i]);
			k = i;
		}
	}
	return P(ma, k);
}

int Move(int start, i64 next_t){
	P choice = choice_next(start, next_t);
	i64 ma = choice.first, k = choice.second;
	if(ma >= para[1]){
		int npos = route[start][k][0];
		for(int i = 0; i < dist[now][npos]; i++){
			if(next_t >= T) break;
			//cout << npos << endl;
			o[Times][next_t] = npos;
			next_t++;
		}
		now = npos;
		for(int i = 0; i < have[now]; i++){
			result[Times] += T*T - (next_t - v[now][finished[now]])*(next_t - v[now][finished[now]]);
			finished[now]++;
		}
		have[now] = 0;
		
		next_t = Move(now, next_t);
		
		return next_t;
	}
	for(int i = route[1][now].size()-2; i >= -1; i = route[1][now].size()-2){
		int npos;
		if(i == -1) npos = 1;
		else npos = route[1][now][i];
		for(int j = 0; j < dist[now][npos]; j++){
			if(next_t >= T) break;
			//cout << npos << endl;
			o[Times][next_t] = npos;
			next_t++;
		}
		now = npos;
	}
	return next_t;
}

int main(){
	cin >> n >> m;
	for(int i = 0; i <= n; i++){
		dist[i][i] = 0;
	}
	for(int i = 0; i < m; i++){
		int a,b,c;
		cin >> a >> b >> c;
		G[a].push_back(P(b,c));
		G[b].push_back(P(a,c));
		dist[a][b] = c;
		dist[b][a] = c;
		route[a][b].push_back(b);
		route[b][a].push_back(a);
	}
	cin >> T;
	for(int i = 0; i < T; i++){
		int a,b,c;
		cin >> a;
		if(a){
			cin >> b >> c;
			v[c].push_back(i);
			to[i] = c;
		}
	}
	
	floyd();
	
	for(int para1 = 1; para1 <= 512; para1*=2) for(int para2 = 1; para2 <= 128; para2*=2){
		para[0] = para1; para[1] = para2;
		result.push_back(0);
		for(int i = 0; i <= n; i++){
			have[i] = 0;
		}
		now = 1; next_t = 0;
		for(i64 t = 0; t < T; t++){
			if(now == 1) have[to[t]]++;
			if(next_t > t) continue;
			next_t = Move(1, t);
			if(next_t == t){
				//cout << -1 << endl;
				o[Times][next_t] = -1;
			}
		}
		Times++;
	}
	
	i64 ans_cnt = 0, ans_time = 0;
	for(int i = 0; i < Times; i++){
		if(ans_cnt < result[i]){
			ans_cnt = result[i];
			ans_time = i;
		}
	}
	for(int i = 0; i < T; i++){
		cout << o[ans_time][i] << endl;
	}
	
	return 0;
}