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,n) for (int i=0;i<(n);i++)
#define REP2(i,m,n) for (int i=m;i<(n);i++)
typedef long long ll;

const int INF = 1 << 29;

int N, M, Tmax;
int G[400][400];
int order_dest[10000];
int dist[400][400];

int current_dest = 0, remaining_dist = 0, latest_order = -1;
int command[10000];

bool is_at_store() {
    return current_dest == 0 && remaining_dist == 0;
}

void search() {
    vector<int> dests;
    REP(t, Tmax) {

        if (remaining_dist > 0) {
            command[t] = current_dest;
            remaining_dist -= 1;
        }

        if (remaining_dist > 0) {
            continue;
        }

        if (is_at_store()) {
            for (int i = 0; i < dests.size(); ) {
                if (dests[i] == 0) {
                    swap(dests[i], dests[(int)dests.size()-1]);
                    dests.pop_back();
                } else {
                    i += 1;
                }
            }
            while (latest_order < t) {
                latest_order += 1;
                if (order_dest[latest_order] != 0) {
                    dests.push_back(order_dest[latest_order]);
                }
            }
        }

        if (is_at_store() && dests.empty())  {
            command[t] = -1;
            continue;
        }

        for (int i = 0; i < dests.size(); ) {
            if (dests[i] == current_dest) {
                swap(dests[i], dests[(int)dests.size()-1]);
                dests.pop_back();
            } else {
                i += 1;
            }
        }

        if (dests.empty()) {
            dests.push_back(0);
        }

        int mdest = -1;
        int mdist = INF;

        for (auto dest: dests) {
            int d = dist[current_dest][dest];
            if (d < mdist) {
                mdist = d;
                mdest = dest;
            }
        }

        int mdest2 = -1;
        int mdist2 = INF;

        REP(i, N) if (i != current_dest && G[current_dest][i] != INF) {
            int d = G[current_dest][i] + dist[i][mdest];
            if (d < mdist2) {
                mdist2 = d;
                mdest2 = i;
            }
        }

        command[t] = mdest2;
        remaining_dist = G[current_dest][mdest2];
        current_dest = mdest2;
    }
}

void warshall_floyd() {
    REP(i, N) REP(j, N) {
        dist[i][j] = G[i][j];
    }
    REP(i, N) REP(j, N) REP(k, N) {
        dist[j][k] = min(dist[j][k], dist[j][i] + dist[i][k]);
    }
}

void input() {
    cin >> N >> M;
    REP(i, N) REP(j, N) {
        G[i][j] = i == j ? 0 : INF;
    }
    REP(i, M) {
        int u, v, d;
        cin >> u >> v >> d;
        --u, --v;
        G[u][v] = d;
        G[v][u] = d;
    }
    cin >> Tmax;
    REP(i, Tmax) {
        int n; cin >> n;
        if (n == 1) {
            cin >> n >> order_dest[i];
            order_dest[i] -= 1;
            //cerr << order_dest[i] << endl;
        }
    }
}

void output() {
    REP(t, Tmax) {
        cout << (command[t] == -1 ? command[t] : command[t] + 1) << "\n";
    }
}

int main() {
    input();
    warshall_floyd();
    search();
    output();
}