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;
 
vector<vector<int>> g;
vector<vector<int>> dist;
vector<vector<vector<int>>> path;
vector<int> ans;
 
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 Move(int & now, int dest, int & now_time){
    for(int i = 1; i < path[now][dest].size(); i++){
        REP(j, g[path[now][dest][i - 1]][path[now][dest][i]]){
            ans.emplace_back(path[now][dest][i] + 1);
            now_time++;
        }
    }
    now = dest;
    return;
}

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;
}
 
signed main(){
    int V,E;
    cin >> V >> E;
    vector<vector<P>> G(V); // 隣接リスト
    g.assign(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);
        g[u][v] = d;
        g[v][u] = d;
    }
    int T;
    cin >> T;
    vector<int> order[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.assign(V, vector<int>(0));
    path.assign(V, vector<vector<int>>(0, vector<int>(0)));
    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
    }

    /*
    auto f = [&dist](int a, int b){ return dist[0][a] > dist[0][b]; };
    priority_queue<int, vector<int>, decltype(f)> que(f); // 持っている商品のうち最も店舗から近いものが上にくる優先度付きキュー
    */
    vector<int> have(V); // 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
    int pre_time = 0, now_time = 0; // now_time : 現在の時刻
    while(ans.size() <= T){
        if(now_time >= T) break;
        for(int i = pre_time; i <= now_time; i++){
            REP(j,order[i].size()){
                have[order[i][j]]++;
            }
        }
        pre_time = now_time + 1;
        int cnt = 0;
        REP(i,V){
            if(have[i] > 0){
                cnt++;
            }
        }
        if(cnt > 1){
            for(int i = 0; i < cnt; i++){
                int dest = Nearest(now, have);
                Move(now, dest, now_time);
                have[dest] = 0;
            }
            Move(now, 0, now_time);
        }else{
            ans.emplace_back(-1);
            now_time++;
        }
    }
 
    // ここから出力
    REP(i,T){
        if(i >= ans.size()) cout << -1 << endl;
        else cout << ans[i] << endl;
    }
    return 0;
}