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>
#define _overload3(_1, _2, _3, name, ...) name
#define _rep(i, n) repi(i, 0, n)
#define repi(i, a, b) for (ll i = (ll)(a); i < (ll)(b); ++i)
#define rep(...) _overload3(__VA_ARGS__, repi, _rep, )(__VA_ARGS__)
#define ll long long
#define lld long double
#define ALL(x) x.begin(), x.end()
#ifdef DEBUG
#define line() cerr << "[" << __LINE__ << "] ";
#define dump(i) cerr << #i ": " << i << " ";
#define dumpl(i) cerr << #i ": " << i << endl;
#else
#define line(i)
#define dump(i)
#define dumpl(i)
#endif
#define INF 1 << 29
using namespace std;
int maxTime;
int v, e;
int curTime = 0;

void warshallfloyd(vector<vector<int>> &g, vector<vector<int>> &cost, vector<vector<vector<int>>> &path)
{
    int v = g.size();
    rep(i, v) rep(j, v)
    {
        cost[i][j] = g[i][j];
        if (cost[i][j] != INF && cost[i][j] != 0)
        {
            path[i][j].push_back(i);
            path[i][j].push_back(j);
        }
    }
    rep(i, v)
    {
        path[i][i].push_back(i);
    }
    rep(k, v) rep(i, v) rep(j, v)
    {
        if (cost[i][j] > cost[i][k] + cost[k][j])
        {
            cost[i][j] = cost[i][k] + cost[k][j];
            path[i][j] = path[i][k];
            path[i][j].insert(path[i][j].end(), path[k][j].begin() + 1, path[k][j].end());
        }
    }
}

void dumpShortestPathInfo(vector<vector<vector<int>>> &path, vector<vector<int>> &cost)
{
    int v = path.size();
    rep(i, v)
    {
        cout << 1 << "->" << i + 1 << " : ";
        for (auto x : path[0][i])
        {
            cout << x + 1 << " ";
        }
        cout << " (" << cost[0][i] << ")" << endl;
    }
}

void deliver(deque<int> &que, vector<vector<vector<int>>> &path, vector<vector<int>> &cost)
{
    int curPos = 0;
    int curTarget;
    map<int, bool> visited;
    while (!que.empty())
    {
        curTarget = que.front();
        que.pop_front();
        if (visited[curTarget])
            continue;
        for (auto x : path[curPos][curTarget])
        {
            rep(i, cost[curPos][x])
            {
                cout << x + 1 << endl;
                curTime++;
                if (curTime == maxTime)
                    exit(0);
            }
            curPos = x;
            visited[curPos] = true;
        }
    }
    for (auto x : path[curPos][0])
    {
        rep(i, cost[curPos][x])
        {
            cout << x + 1 << endl;
            curTime++;
            if (curTime == maxTime)
                exit(0);
        }
        curPos = x;
    }
}

int main()
{
    cin >> v >> e;
    vector<vector<int>> g(v, vector<int>(v, INF));
    vector<vector<int>> cost(v, vector<int>(v, INF));

    vector<vector<vector<int>>> path(v, vector<vector<int>>(v));

    rep(i, e)
    {
        int a, b, c;
        cin >> a >> b >> c;
        a--;
        b--;
        g[a][b] = c;
        g[b][a] = c;
    }
    rep(i, v)
    {
        g[i][i] = 0;
    }
    warshallfloyd(g, cost, path);
    cin >> maxTime;
    vector<pair<int, int>> orders(maxTime, {-1, -1});
    int orderIndex = 0;

    rep(i, maxTime)
    {
        int t;
        cin >> t;
        if (t == 1)
        {
            cin >> orders[orderIndex].first >> orders[orderIndex].second;
            orders[orderIndex].first--;
            orders[orderIndex].second--;
            orderIndex++;
        }
    }
    deque<int> que;
    orderIndex = 0;

    while (true)
    {
        while (orders[orderIndex].first != -1 && orders[orderIndex].first <= curTime)
        {
            que.push_back(orders[orderIndex].second);
            orderIndex++;
        }
        if (que.empty())
        {
            cout << -1 << endl;
            curTime++;
            if (curTime == maxTime)
                exit(0);
        }
        else
        {
            deliver(que, path, cost);
        }
    }
    return 0;
}