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:
// https://atcoder.jp/contests/hokudai-hitachi2019-1/tasks/hokudai_hitachi2019_1_a
#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) FOR(i, 0, n)
#define FOR(i, s, n) for (int i = (s), i##_len = (n); i < i##_len; ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define ALLR(obj) (obj).rbegin(), (obj).rend()
const int INF = 1e9;

template <typename T> using Matrix = vector<vector<T>>;
template <typename T> void warshall_floyd(Matrix<T> &d, Matrix<T> &n) {
    int V = d.size();
    REP(k, V) REP(i, V) REP(j, V) {
        if (d[i][k] + d[k][j] < d[i][j]) {
            d[i][j] = d[i][k] + d[k][j];
            n[i][j] = n[i][k];
        } else if (k != i && d[i][k] + d[k][j] == d[i][j]) {
            n[i][j] = min(n[i][j], n[i][k]);
        }
    }
}

int V, E, T;

vector<int> find_route(const Matrix<int> &mat, const Matrix<int> &nex,
                       const Matrix<int> &in, int freq) {
    int now_pos = 0, count = 0, start = T * 0.9, LT = T;
    vector<int> ans, mp;
    // 関数化してans.size()が大きいルートを採用
    unordered_set<int> ha;
    queue<int> q;
    REP(i, T) {
        LT--;
        mp = in[i];
        // 開始位置なら荷物を乗せる
        if (now_pos == 0) {
            for (auto &x : mp) {
                ha.insert(x);
            }
            mp.clear();
        }
        if (count == 0) {
            if (!q.empty()) {
                int next_pos = q.front();
                q.pop();
                ans.emplace_back(next_pos);
                count = mat[now_pos][next_pos] - 1;
                now_pos = next_pos;
            } else {
                int next_pos = 0, next_dist = INF;
                // 持っている数より店舗の方がfreq倍多ければ戻る
                if (LT > start || mp.size() > ha.size() * freq) {
                    // 店舗に戻る
                    next_pos = 0;
                } else {
                    // 現在から最寄りの配達先を指定
                    for (auto &x : ha) {
                        if (next_dist > mat[now_pos][x]) {
                            next_dist = mat[now_pos][x];
                            next_pos = x;
                        }
                    }
                    if (next_dist * 2 <= LT) {
                        // 荷物を降ろしたことにする
                        ha.erase(next_pos);
                    } else {
                        next_pos = 0;
                    }
                }
                if (next_pos == now_pos) {
                    // 初期位置なら待機
                    ans.emplace_back(-2);
                } else {
                    for (int cur = nex[now_pos][next_pos]; cur != next_pos;
                         cur = nex[cur][next_pos]) {
                        q.push(cur);
                    }
                    q.push(next_pos);
                    next_pos = q.front();
                    q.pop();
                    ans.emplace_back(next_pos);
                    count = mat[now_pos][next_pos] - 1;
                    now_pos = next_pos;
                }
            }
        } else {
            // 直前と同じ方向へ動く
            ans.emplace_back(ans.back());
            count--;
        }
    }
    return ans;
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> V >> E;
    Matrix<int> mat(V, vector<int>(V, INF)), nex(V, vector<int>(V));
    REP(i, V) {
        mat[i][i] = 0;
        REP(j, V) nex[i][j] = j;
    }

    for (int i = 0; i < E; i++) {
        int x, y, z;
        cin >> x >> y >> z;
        x--, y--;
        mat[x][y] = z;
        mat[y][x] = z;
    }
    warshall_floyd(mat, nex);
    cin >> T;
    vector<int> ans;
    Matrix<int> in(T);
    REP(i, T) {
        int N;
        cin >> N;
        while (N--) {
            int id, dst;
            cin >> id >> dst;
            dst--;
            in[i].emplace_back(dst);
        }
    }
    // 所持している荷物 * i < 店舗の注文なら店舗に戻る
    int maxcnt = 0;
    FOR(i, 1, 25) {
        vector<int> cmd = find_route(mat, nex, in, i);
        // -2以外の数を数える
        int cnt = 0;
        for (auto &x : cmd) {
            if (x != -2) {
                cnt++;
            }
            if (cnt > maxcnt) {
                ans = cmd;
            }
        }
    }

    // 3-opt 法と 2-opt 法を同時に用いる
    // https://www.cst.nihon-u.ac.jp/research/gakujutu/57/pdf/L-20.pdf

    // 出力する
    int now_pos = 0, cnt = 0;
    for (auto &x : ans) {
        if (x == -2) {
            cout << "-1\n";
            if (++cnt >= T)
                break;
        } else {
            int next_dist = mat[now_pos][x];
            REP(i, next_dist) {
                cout << x + 1 << "\n";
                if (++cnt >= T)
                    break;
            }
            now_pos = x;
        }
    }
    return 0;
}