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 <fstream>
#include <numeric>
#include <stdio.h>
#include <cmath>
#include <queue>
#include <iomanip>
#include <algorithm>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <deque>
#include <bitset>
#include <cstring>
#include <sstream>
 
#define f first
#define s second
#define ll long long
#define ld long double
#define pb push_back
#define all(x) x.begin(),x.end()
#define mp make_pair
#define y0 eto
#define y1 mezhdy
#define y2 nami
#define left extermination
#define right dismemberment
 
using namespace std;
 
typedef pair<int,int> pii;
 
template <class T>void read(vector<T> &a,ll n){T x; a.clear();for(ll i=0;i<n;i++){cin>>x;a.pb(x);}}
template <class T>void write(vector<T> &a){for(T x : a) cout<<x<<" ";cout<<endl;}

const int N=405;

int n,m,Tmax;
int p[N];

map<pii,int> have;
vector<pair<int,pii> > orders;
vector<pair<int,pii> > e1;
int last_w;

int tmm;

vector<pii> g[N];
int d[N];
int par[N];

vector<int> ans;

void prepare_dsu()
{
    for(int i=0;i<N;i++)
        p[i]=i;
}

int fnd(int u)
{
    if(u==p[u]) return u;
    return p[u]=fnd(p[u]);
}

void un(int a,int b)
{
    int u=fnd(a);
    int v=fnd(b);
    if(rand()&1) swap(u,v);
    if(u!=v)
        p[u]=v;
}


void refresh(int u)
{
    vector<pii> v;
    while(par[u]!=-1)
    {
        v.pb({u,par[u]});
        u=par[u];
    }
    
    for(int i=0;i<v.size();i++)
    {
        int a=v[i].f;
        int b=v[i].s;
        for(int j=0;j<have[{a,b}];j++)
            ans.pb(b);
    }
    
    for(int i=0;i<v.size();i++)
        swap(v[i].f,v[i].s);
    reverse(all(v));
    
    for(int i=0;i<v.size();i++)
    {
        int a=v[i].f;
        int b=v[i].s;
        for(int j=0;j<have[{a,b}];j++)
            ans.pb(b);
    }
}

void go1(int u,int p=-1)
{
    if(ans.size()>Tmax) return;
    
    tmm++;
    if(u==1)
        tmm=0;
    
    if(tmm>100)
    {
        refresh(u);
        tmm=0;
    }
    
    for(int i=0;i<g[u].size();i++)
    {
        if(g[u][i].f==p) continue;
        
        for(int j=0;j<g[u][i].s;j++)
            ans.pb(g[u][i].f);
        
        go1(g[u][i].f,u);
        
        for(int j=0;j<g[u][i].s;j++)
            ans.pb(u);
    }
    
}



void dfs(int u,int p=-1,int dst=0)
{
    d[u]=dst;
    par[u]=p;
    for(int i=0;i<g[u].size();i++)
        if(g[u][i].f!=p)
            dfs(g[u][i].f,u,dst+1);
}

void ride1()
{
    go1(1);
}


int main()
{
    ios_base::sync_with_stdio(0);
    //freopen("scripts/test.in", "r", stdin);
    //freopen("scripts/result.out", "w", stdout);
    
    cin>>n>>m;
    for(int i=0;i<m;i++)
    {
        int a,b,c;
        cin>>a>>b>>c;
        e1.pb(mp(c,mp(a,b)));
        have[{a,b}]=c;
    }
    
    cin>>Tmax;
    for(int i=0;i<Tmax;i++)
    {
        int Nnew;
        cin>>Nnew;
        while(Nnew--)
        {
            int id,x;
            cin>>id>>x;
            orders.pb(mp(i,mp(id,x)));
        }
    }
    prepare_dsu();
    sort(all(e1));
    
    for(int i=0;i<e1.size();i++)
    {
        int a=e1[i].s.f;
        int b=e1[i].s.s;
        int w=e1[i].f;
        if(fnd(a)!=fnd(b))
        {
            un(a,b);
            g[a].pb(mp(b,w));
            g[b].pb(mp(a,w));
            have[{a,b}]=w;
            have[{b,a}]=w;
        }
    }
    
    dfs(1);
    
    while(ans.size()<Tmax)
        ride1();
    
    while(ans.size()>Tmax)
        ans.pop_back();
    while(ans.size()<Tmax)
        ans.pb(-1);
    
    for(int i=0;i<ans.size();i++)
        cout<<ans[i]<<endl;
    
}