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;
#define REP(i,a) for(int i = 0; i < (a); i++)
#define ALL(a) (a).begin(),(a).end()
typedef long long ll;
typedef pair<int, int> P;
const int INF = 1e9;
const long long LINF = 1e18;
const long long MOD = 1e9 + 7;
 
int T;
vector<vector<int>> edge_length;
vector<vector<int>> order;
vector<vector<int>> dist;
vector<vector<vector<int>>> path;
vector<int> have;
int pre_time = 0, now_time = 0; // now_time : 現在の時刻
 
template <typename T>
vector<T> Dijkstra(int s, vector<vector<pair<int, T>>> & G, vector<int> & prev, const T INF = 1e9){
    using P = pair<T, int>;
    int V = G.size();
    vector<T> dist(V, INF);
    priority_queue<P, vector<P>, greater<P>> que;
    dist[s] = 0;
    que.emplace(0, s);
    while(!que.empty()){
        P p = que.top();
        que.pop();
        int v = p.second;
        if(dist[v] < p.first) continue;
        for(int i = 0; i < G[v].size(); i++){
            int to = G[v][i].first;
            T cost = G[v][i].second;
            if(dist[to] > dist[v] + cost){
                dist[to] = dist[v] + cost;
                prev[to] = v;
                que.emplace(dist[to], to);
            }
        }
    }
    return dist;
}
 
vector<vector<int>> GetPath(vector<int> & prev){
    int V = prev.size();
    vector<vector<int>> path(V);
    for(int i = 0; i < V; i++){
        for(int t = i; t != -1; t = prev[t]){
            path[i].emplace_back(t);
        }
        reverse(path[i].begin(), path[i].end());
    }
    return path;
}

void Output(int ans){
    if(ans < -1 || ans == 0) ans = -1;
    cout << ans << endl;
    now_time++;
    if(now_time >= T) exit(0);
}
 
void Move(int & now, int dest){
    for(int i = 1; i < path[now][dest].size(); i++){
        REP(j, edge_length[path[now][dest][i - 1]][path[now][dest][i]]){
            Output(path[now][dest][i] + 1);
        }
    }
    now = dest;
}
 
int Nearest(int now, vector<int> & have){
    int nearest, mn = INF;
    REP(i,have.size()){
        if(have[i] > 0 && dist[now][i] < mn){
            mn = dist[now][i];
            nearest = i;
        }
    }
    return nearest;
}

vector<int> Cand(vector<int> & have){
    vector<int> res;
    REP(i,have.size()){
        if(have[i] > 0){
            res.emplace_back(i);
        }
    }
    sort(ALL(res));
    return res;
}

vector<int> Order(vector<int> vec){
    int mn = INF;
    int n = vec.size();
    vector<int> res(n);
    do{
        int sum = 0;
        sum += dist[0][vec[0]];
        for(int i = 0; i < n - 1; i++){
            sum += dist[vec[i]][vec[i + 1]];
        }
        sum += dist[vec[n - 1]][0];
        if(sum < mn){
            mn = sum;
            for(int i = 0; i < n; i++){
                res[i] = vec[i];
            }
        }
    }while(next_permutation(ALL(vec)));
    return res;
}

void Load(){
    for(int i = pre_time; i <= now_time; i++){
        REP(j,order[i].size()){
            have[order[i][j]]++;
        }
    }
    pre_time = now_time + 1;
}
 
signed main(){
    int V,E;
    cin >> V >> E;
    vector<vector<P>> G(V); // 隣接リスト
    edge_length.resize(V, vector<int>(V, INF)); // 隣接行列
    REP(i,E){
        int u,v,d;
        cin >> u >> v >> d;
        u--;
        v--;
        G[u].emplace_back(v, d);
        G[v].emplace_back(u, d);
        edge_length[u][v] = d;
        edge_length[v][u] = d;
    }
    cin >> T;
    order.resize(T); // order[i] : 時刻 i に発生する注文
    REP(i,T){
        int n;
        cin >> n;
        REP(j,n){
            int id,dst;
            cin >> id >> dst;
            dst--;
            order[i].emplace_back(dst);
        }
    }
    // ここまで入力
 
    vector<vector<int>> prev(V, vector<int>(V, -1));
    dist.resize(V);
    path.resize(V);
    for(int i = 0; i < V; i++){
        dist[i] = Dijkstra(i, G, prev[i], INF); // 全点間最短経路を求める
        path[i] = GetPath(prev[i]); // 経路復元
    }

    have.resize(V, 0); // have[i] : 車に積んである頂点 i への商品の数
    int now = 0; // now : 現在地
    while(true){
        Load();
        int cnt = 0;
        REP(i,V){
            if(have[i] > 0){
                cnt++;
            }
        }
        if(cnt > 2){
            
            if(cnt < 11){
                vector<int> tmp = Cand(have);
                vector<int> vec = Order(tmp);
                for(int i = 0; i < vec.size(); i++){
                    Move(now, vec[i]);
                    have[vec[i]] = 0;
                }
            }else{
                for(int i = 0; i < cnt; i++){
                    int dest = Nearest(now, have);
                    Move(now, dest);
                    have[dest] = 0;
                }
            }
            /*
            for(int i = 0; i < cnt; i++){
                int dest = Nearest(now, have);
                Move(now, dest);
                have[dest] = 0;
            }
            */
            Move(now, 0);
        }else{
            Output(-1);
        }
    }
    return 0;
}