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
#define INF 1000000
typedef long long ll;
typedef pair<int,int> pii;//  t,dst_idx

struct edge {
    int to;
    int cost;
};
struct order {
    int dst;
    int id;
    int issued_time;

    bool operator<(order other) const
    {
        return issued_time < other.issued_time;
    }
};
int d[VERTEX_MAX][VERTEX_MAX]; //d[s][d] -> s,d間の最短距離を持つ(ワーシャルフロイド)
int warshall_next[VERTEX_MAX][VERTEX_MAX]; // warshall_next[i][j] -> i-jの最短路におけるiの次の点
void warshall_floyd(){
    REP(k, VERTEX_MAX) {
        REP(i, VERTEX_MAX) {
            REP(j, VERTEX_MAX) {
                int pre_dij = d[i][j];
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
                if (pre_dij != d[i][j]){ // 同じなら少しでも多く経由したい
                    warshall_next[i][j] = warshall_next[i][k];
                }
            }
        }
    }
}

int dst2op(int src, int dst){// 最短でsrcからdstに行くときのsrcの次に経由するノードを返す
    if (src == dst ) return -1;
    return warshall_next[src][dst];
}

bool kaeritai(int nearest, int rest, int before_load, int car_pos) {
    if(before_load == 0) return false;
    if(nearest > d[car_pos][1]*3 || rest * 4 < before_load ) {
        return true;
    }
    return false;
}

vector<vector<edge>> G(VERTEX_MAX);
bool Tlast = false;
int main(int argc, char const *argv[])
{
    cin.tie(0);
   	ios::sync_with_stdio(false);
    vector<order> v_O;// dst, issued_time
    // ワーシャルフロイド法初期化
    REP(i, VERTEX_MAX) {
        REP(j, VERTEX_MAX) {
            d[i][j] = INF;
            if (i == j && i != 0) {
                d[i][j] = 0;
            }
            warshall_next[i][j] = j;
        }
    }
    //handle input
    int num_V, num_E; cin >> num_V >> num_E;
    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);
        d[tmp_u][tmp_v] = tmp_d;
        d[tmp_v][tmp_u] = tmp_d;
    }
    int T; cin >> T;
    REP(t, T) {
        int num_O; cin >> num_O;
        REP(i, num_O) {
            order tmp; cin >> tmp.id >> tmp.dst;
            tmp.issued_time = t;
            v_O.push_back(tmp);
        }
    }
    //全点対最短経路
    warshall_floyd();
    // handle status
    int car_pos = 1; // vertex_idx
    // orderを時間昇順に
    sort(v_O.begin(),v_O.end());
    //handle steps
    int op = -1;
    auto already_load = v_O.begin();
    vector<order> v_loads;
    int final_goal = -1;
    
    REP(t, T) {
        if (car_pos == 1) { // 店にいる時(たまたま通過しうる)
            order comp_order = {VERTEX_MAX, 0, t};
            auto res = upper_bound(v_O.begin(), v_O.end(), comp_order);
            // 積み込みa
            for (auto itr = already_load; itr != res ;itr++){
                order o;
                o.issued_time = itr->issued_time;
                o.dst = itr->dst;
                o.id = itr->id;
                v_loads.push_back(o);
            }
            // どこに行くか判定
            int min_d = INF;
            int min_dst = 1;
            int max_wating = 0;
            int max_dst = 1;
            for (auto o: v_loads) {
                min_d = min(min_d, d[1][o.dst]);
                if (min_d == d[1][o.dst]) min_dst = o.dst;
                max_wating = max(max_wating, t-o.issued_time);
                if (max_wating == t-o.issued_time) max_dst = o.dst;
            }
            final_goal = min_dst;
            if (max_wating > 500) final_goal = max_dst;//(めっちゃ待つ人をいつ回収すべきか？)
            op = dst2op(car_pos ,final_goal);
            already_load = res;
        }
        else if (car_pos > 0) {// 店以外の頂点にいる時
            // 荷降ろし
            v_loads.erase(remove_if(v_loads.begin(),v_loads.end(), [&car_pos](order o) { return o.dst == car_pos; }),v_loads.end());
            order comp_order = {VERTEX_MAX, 0, t};
            auto res = upper_bound(v_O.begin(), v_O.end(), comp_order);
            int before_put = distance(already_load,res);
            //どこ行くか判定
            if (final_goal == car_pos) {
                //ここを基本的には近いゴールに配達し、帰巣本能が強くなったら帰る感じにする
                int min_d = INF;
                int min_dst = 1;
                for (auto o: v_loads) {
                    min_d = min(min_d, d[car_pos][o.dst]);
                    if (min_d == d[car_pos][o.dst]) min_dst = o.dst;
                }
                final_goal = min_dst;
                if (kaeritai(min_d, (int)v_loads.size(), before_put, car_pos)) final_goal = 1;
                if (!Tlast && T*0.95 < t + d[car_pos][1]) {
                    Tlast = true;
                    final_goal = 1;
                }
                op = dst2op(car_pos,final_goal);
            }
            else {
                op = dst2op(car_pos,final_goal);
            }
        }
        
        if (op == -1) {
            cout << op << endl;
        }
        else { // opの示すノードまで進む
            for(int t_next = t + d[car_pos][op]; t<T && t<t_next ;t++) {
                cout << op << endl;
            }
            t--;
            car_pos = op;
        }
    }
    return 0;
}