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 <queue>
#include <algorithm>
#include <vector>
#include <math.h>
using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;

int V, E, INF = 1e9, d[500][500], d2[500][500];
int Tmax, nxt[500][500];
pair<int, int> order[10010];

ll score_calc(vector<int> output) {
    ll score = 0;
    int from = 1, to = -1, at = 0, id = 0;
    vector<int> dst[500];
    for (int i = 0; i < Tmax; i++) {
        if (order[id].first == i) {
            dst[order[id].second + 1].push_back(i);
            id++;
        }
        if (output[i] != -1) {
            if (to == -1) {
                to = output[i];
                at = 1;
            }
            else {
                if (to == output[i]) {
                    at++;
                    if (at == d2[from - 1][to - 1]) {
                        from = to;
                        to = -1;
                        at = 0;

                        for (int j = 0; j < dst[from].size(); j++) {
                            score += Tmax * Tmax - pow(i - dst[from][j], 2);
                        }
                        dst[from].clear();
                    }
                }
                else {
                    //cout << "out: " << i  << " " << output[i] << " " << from << " " << to << " " << at << " ";
                    //cout << d2[from - 1][to - 1] << " ";
                    return -1;
                }
            }
        }
    }
    return score;
}

void warshall_floyd() {
    for (int i = 0; i < V; i++) {
        for (int j = 0; j < V; j++) nxt[i][j] = j;
    }
    for (int k = 0; k < V; k++) {
        for (int i = 0; i < V; i++) {
            for (int j = 0; j < V; j++) {
                if (d2[i][k] + d2[k][j] < d2[i][j]) {
                    d2[i][j] = min(d2[i][j], d2[i][k] + d2[k][j]);
                    nxt[i][j] = nxt[i][k];
                }
            }
        }
    }        
}

vector<int> shortestPath(int start, int goal) {
    vector<int> path;
    for (int cur = start; cur != goal; cur = nxt[cur][goal]) {
        path.push_back(cur);
    }
    path.push_back(goal);
    return path;
}

int main() {
    
    cin >> V >> E;

    for (int i = 0; i < V; i++) {
        for (int j = 0; j < V; j++) {
            d[i][j] = INF;
        }
    }
    
    for (int i = 0; i < E; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        u--;  v--;
        d[u][v] = w;
        d[v][u] = w;
    }
    
    for (int i = 0; i < V; i++) {
        for (int j = 0; j < V; j++) {
            d2[i][j] = d[i][j];
        }
    }

    warshall_floyd();

    
    cin >> Tmax;
    for (int i = 0; i < Tmax; i++) {
        int Nnew;
        cin >> Nnew;
        if (Nnew == 1) {
            int ID, dst;
            cin >> ID >> dst;
            order[ID - 1] = make_pair(i, dst - 1);
        }
    }

    int id = 0, i = 0;
    vector<int> output;
    while (i < Tmax) {
        if (i < order[id].first) {
            output.push_back(-1);
            i++;
        }
        else {
            int tmp = 0;
            vector<int> route = shortestPath(0, order[id].second);
            for (int j = 0; j < route.size() - 1; j++) {
                for (int k = 0; k < d[route[j]][route[j + 1]]; k++) {
                    output.push_back(route[j + 1] + 1);
                    tmp++;
                }
            }
            route = shortestPath(order[id].second, 0);
            for (int j = 0; j < route.size() - 1; j++) {
                for (int k = 0; k < d[route[j]][route[j + 1]]; k++) {
                    output.push_back(route[j + 1] + 1);
                    tmp++;
                }
            }
            i += tmp;
            id++;
        }
    }

    //cout << "score: " << score_calc(output) << endl;
    for (int i = 0; i < Tmax; i++) cout << output[i] << endl;
    /*for (int i = 0; i < V; i++) {
        for (int j = 0; j < V; j++) {
            if (d[i][j] == INF) cout << -1 << " ";
            else cout << d[i][j] << " ";
        }
        cout << endl;
    }
    //*/
}