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 VERTEX_MAX 401

typedef long long ll;
typedef pair<int,int> pii;//  t,dst_idx
// int G[VERTEX_MAX][VERTEX_MAX];
struct edge {
    int to;
    int cost;
};
vector<vector<edge>> G(VERTEX_MAX);
int main(int argc, char const *argv[])
{
    cin.tie(0);
   	ios::sync_with_stdio(false);
    int num_V, num_E; cin >> num_V >> num_E;
    vector<pii> v_O;// dst, issued_time
    REP(i, num_E) {
        int tmp_u, tmp_v, tmp_d; cin >> tmp_u >> tmp_v >> tmp_d;
        edge tmp1, tmp2;
        tmp1.to = tmp_u; tmp1.cost = tmp_d;
        tmp2.to = tmp_v; tmp2.cost = tmp_d;
        G[tmp_u].push_back(tmp2);
        G[tmp_v].push_back(tmp1);
    }
    int T; cin >> T;
    REP(t, T) {
        int num_O; cin >> num_O;
        REP(i, num_O) {
            int tmp_id, tmp_dst; cin >> tmp_id >> tmp_dst;
            v_O.push_back(make_pair(t,tmp_dst));
        }
    }
    //handle steps
    int car_pos = 1;
    sort(v_O.begin(),v_O.end(),[](const pair<int,int>& lhs, const pair<int,int>& rhs){
        if (lhs.first == rhs.first) return lhs.second < rhs.second;
        else return lhs.first < rhs.first;
    });
    // 1の隣からの注文のみ対応する
    int leave = -1;
    int to_dst = 0;
    REP(t, T) {
        auto res = lower_bound(v_O.begin(), v_O.end(), make_pair(t,VERTEX_MAX), [](const pair<int,int>& lhs, const pair<int,int>& rhs){
            if (lhs.first == rhs.first) return lhs.second < rhs.second;
            else return lhs.first < rhs.first;
        });
        if (leave == -1) {
            for (auto itr = v_O.begin(); itr != res ;itr++){
                int target_dst = itr->second;
                for(auto e: G[1]) {// 店舗からダイレクトにつながっているところだけ
                    if (e.to == target_dst) {
                        leave = target_dst;
                        to_dst = e.cost;
                    }
                }
            }
        }
        if (leave == -1 || leave == 0) {
            cout << -1 << endl;
        }
        else {
            if (to_dst == 0) {
                leave = 0;
                cout << -1 << endl;
            }
            else {
                cout << leave << endl;
                to_dst--;
            }
        }
    }
    return 0;
}