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;
using ll=long long;
const ll MOD=(ll)1e9+7;
const ll INF=(ll)1e18+10;

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

vector<pair<int,int>> tonari[410]; //(next node,dist)
int dist[410][410],to[410][410];
int tmax;

class Order{
public:
    int dst,ot; //目的地と注文時刻 destination ordertime
    bool arrive=false;

    Order(int x,int y){
        dst=y;
        ot=x;
    }
    
    Order(const Order& src){ //コピーコンストラクタ
        dst=src.dst;
        ot=src.ot;
    }
};

vector<int> path(int s, int g){
    vector<int> tmp;
    int pos;
    for (pos = s; pos!=g ; pos=to[pos][g]) {
        tmp.push_back(pos);
    }
    tmp.push_back(g);
    return tmp;
}

int go(int start,int goal,int starttime){ //到着時の時刻を返す
    vector<int> pathtod(path(start,goal));
    //for(int a:pathtod) cout<<a<<" ";
    //cout<<endl<<dist[0][1]<<endl;
    int pos=start,i=1;
    while(pos!=goal&&starttime<tmax){
        //cout<<"syoyo:"<<dist[pos][pathtod[i]]<<endl;
        int tmp=tmax-starttime;
        for (int j = 0; j < min(dist[pos][pathtod[i]], tmp); ++j) {
            //cout<<start<<","<<starttime<<":";
            cout<<pathtod[i]+1<<endl;
            ++starttime;
        }
        pos=pathtod[i];
        ++i;
        //cout<<pos<<endl;
    }
    //cout<<"OK"<<endl;
    return starttime;
}

int main() {
    int N,M; cin>>N>>M;
    for (int i = 0; i < N; ++i) {
        for (int j = 0; j < N; ++j) {
            to[i][j]=j;
            if(i==j) dist[i][j]=0;
            else dist[i][j]=MOD; //初期化
        }
    }
    int u,v,d;
    for (int i = 0; i < M; ++i) {
        cin>>u>>v>>d; --u; --v;
        tonari[u].push_back(make_pair(v,d));
        tonari[v].push_back(make_pair(u,d));
        dist[u][v]=d;
        dist[v][u]=d;
    }

    //ワーシャルフロイド
    for (int k = 0; k < N; ++k) {
        for (int i = 0; i < N; ++i) {
            for (int j = 0; j < N; ++j) {
                if(chmin(dist[i][j],dist[i][k]+dist[k][j])){
                    to[i][j]=to[i][k];
                }
            }
        }
    }

    cin>>tmax;
    //注文情報を受け取る
    int Ncnt,id,dst;
    vector<Order> od;
    for (int i = 0; i < tmax; ++i) {
        cin>>Ncnt;
        if(Ncnt){
            cin>>id>>dst; --dst;
            od.push_back(Order(i,dst));
        }
    }

    for(auto a:od){
        //cout<<a.ot<<" "<<a.dst<<endl;
    }
    
    //まずは注文順で毎回店に戻る
    int prenode=0,dstnode=0;
    int update=0; //最後にノード0に来た時刻
    int i=0,t=0;
    vector<int> nextpath;

    while(t<tmax){
        if(t<od[0].ot||(prenode==0&&t<od[i].ot)||i>=od.size()) {
            cout<<-1<<endl;
            ++t;
            continue; //最初の注文までは動かない
        }

        if(prenode==0){
            dstnode=od[i].dst;
            t=go(prenode,dstnode,t);
            prenode=dstnode;
            ++i;
        }else{
            dstnode=0;
            t=go(prenode,dstnode,t);
            prenode=0;
        }
    }


    return 0;
}