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:
/**
 * Libraries
 */
#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
#include <string>
#include <queue>
#include <iomanip>
#include <set>
#include <unordered_map>

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

using namespace std;
using lli = long long int;

const int INF = 1000000007;
const int TOO_MANY_ORDERS = 10;

struct order {
    int id, dstV;
    int time;
    bool operator<(const order x) const {
        return this->dstV < x.dstV;
    }
};

int dist[400][400];
int nextPath[400][400]; // 経路復元用

int T;

vector<int> dstList;
vector<order> orderList; // 順序は保証できないかも
vector<order> waitingList;

lli calcScore(int t, int car, const order& o) {
    lli deliveredTime = dist[car][o.dstV] + t;
    lli waitingTime = deliveredTime - o.time;
    return T * T - waitingTime * waitingTime;
}

int getNextDst(int t, int car) {
    int orderNum = orderList.size();
    if (orderNum == 0) {
        return 0;
    } else if (orderNum >= TOO_MANY_ORDERS) {
        order maxO = orderList[0];
        for (order o: orderList) {
            if (calcScore(t, car, o) > calcScore(t, car, maxO)) {
                maxO = o;
            }
        }

        return maxO.dstV;
    }

    // orderNumが大きくなければ、全探索してスコア最大化を目指す
    order nextO;
    lli maxScore = -1 * INF;
    sort(orderList.begin(), orderList.end());
    do {
        lli score = 0l;
        for (order o: orderList) {
            score += calcScore(t, car, o);
            t += dist[car][o.dstV];
            car = o.dstV;
        }
        // cout << "score...: " << score << "maxScore..." << maxScore << endl;
        if (maxScore < score) {
            maxScore = score;
            nextO = orderList[0];
        }
    } while (next_permutation(orderList.begin(), orderList.end()));

    if (waitingList.size() != 0) {
        order maxWO = waitingList[0];
        for (order o: waitingList) {
            if (calcScore(t + dist[car][0], 0, o) > calcScore(t + dist[car][0], 0, maxWO)) {
                maxWO = o;
            }
        }
        if (calcScore(t + dist[car][0], 0, maxWO) >= calcScore(t, car, nextO)) {
            return 0;
        }
    }
    return nextO.dstV;
}

signed main() {
    int V, E; cin >> V >> E;
    for (int i = 0; i < V; i++) {
        for (int j = 0; j < V; j++) {
            dist[i][j] = i == j ? 0 : INF;
            nextPath[i][j] = j;
        }
    }
    for (int i = 0; i < E; i++) {
        int u, v, d; cin >> u >> v >> d; u--; v--;
        dist[u][v] = d;
        dist[v][u] = d;
    }
    for (int k = 0; k < V; k++) {
        for (int i = 0; i < V; i++) {
            for (int j = 0; j < V; j++) {
                if (dist[i][j] > dist[i][k] + dist[k][j]) {
                    dist[i][j] = dist[i][k] + dist[k][j];
                    nextPath[i][j] = nextPath[i][k];
                }
            }
        }
    }

    cin >> T;
    int car = 0;
    int skipTo = 0;
    int dstV = 0, gotoV = 0;
    // vector<int> cargoInStore(T, -1);
    // vector<int> cargoOnCar;
    for (int t = 0; t < T; t++) {
        // cout << "time: " << t << endl;
        int N; cin >> N;
        vector<order> orders(N);
        for (order o: orders) {
            cin >> o.id >> o.dstV; o.dstV--; o.time = t;
            waitingList.push_back(o);
        }

        // 辺上を移動する
        // cout << "move on edge: " << t << endl;
        if (t < skipTo) {
            cout << gotoV + 1 << endl; // move w
            continue;
        } else if (skipTo == t) {
            skipTo = -1;
            car = gotoV;
        }

        // 積荷回収
        // cout << "salvage: " << t << endl;
        if (car == 0) {
            for (order o: waitingList) {
                orderList.push_back(o);
            }
            waitingList.clear();
        } else if (car == dstV) {
            auto result = remove_if(orderList.begin(), orderList.end(),
                    [=](order o) { return o.dstV == car; });
            orderList.erase(result, orderList.end());
        }

        // 目的地更新
        // cout << "arrive[having orders: " << orderList.size() << "]: " << t << endl;
        dstV = getNextDst(t, car);

        // 短期目的地更新
        // cout << "get destination(" << dstV << "): " << t << endl;
        gotoV = nextPath[car][dstV];
        skipTo = t + dist[car][gotoV];

        // 意思決定
        // cout << "detect command: " << t << endl;
        if (gotoV == car) {
            cout << -1 << endl; // stay
        } else {
            cout << gotoV + 1 << endl; // move w
        }
    }
    return 0;
}