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 PYTHON3):

Solution:
import sys
from collections import deque
from heapq import heapify, heappop, heappush

def decide_next_dst(cost, order):
    d = 0
    c = 0
    for key in order:
        if cost[key] > c:
            d = key
            c = cost[key]
    return (d, c)

def solve():
    inf = 10000000000
    input = sys.stdin.readline
    
    V, E = map(int, input().split())
    edge = dict()
    wf = dict()
    next_id = dict() #iからjに行くときにiの次に訪れる地点
    for i in range(V):
        edge[i] = dict()
        wf[i] = dict()
        next_id[i] = dict()
        for j in range(V):
            edge[i][j] = (0 if i == j else inf)
            next_id[i][j] = j
            wf[i][j] = (0 if i == j else inf)

    for _ in range(E):
        u, v, d = map(int, input().split())
        edge[u - 1][v - 1] = d
        edge[v - 1][u - 1] = d
        wf[u - 1][v - 1] = d
        wf[v - 1][u - 1] = d

    #全頂点最短経路と目的の頂点に向かうとき次に行くべき頂点の復元
    for k in range(V): 
        for i in range(V):
            for j in range(V):
                if wf[i][j] > wf[i][k] + wf[k][j]:
                    wf[i][j] = wf[i][k] + wf[k][j]
                    next_id[i][j] = next_id[i][k] 

    T = int(input())
    order = set()
    stuck_order = set()

    command = [None] * T
    heading = 0 #次に向かう地点
    dist_left = 0 #次に向かう地点までの残り距離
    final_dist = 0
    stuck_cost = [0 for _ in range(V)] 
    cost = [0 for _ in range(V)] 
    driver_hold = 0
    store_hold = 0

    for t in range(T):
        N = int(input()) #注文の発生
        if N == 1:
            new_id, dst = map(int, input().split())
            stuck_order |= {dst - 1}
            stuck_cost[dst - 1] += 1
            store_hold += 1

        if dist_left > 0: #移動中の場合そのまま移動を続ける
            command[t] = heading + 1
            dist_left -= 1
        
        else:
            if heading == 0: #店にいるときの処理
                if store_hold == driver_hold == 0: 
                    command[t] = -1
                    continue
                else:
                    order |= stuck_order
                    for key in order: 
                        cost[key] += stuck_cost[key]
                        stuck_cost[key] = 0
                    driver_hold = sum(cost)
                    stuck_order = set()
                    store_hold = 0
        
            if heading in order and heading > 0: #顧客のいる場所で荷物を積み下ろすとき
                order -= {heading} 
                driver_hold -= cost[heading]
                cost[heading] = 0

            current_id = heading #現在地の更新
            if len(order) > 0: #まだ配達すべき荷物があるとき
                if current_id == final_dist: #目的地に到着したときは残りの荷物で先に運ぶべき荷物を選ぶ
                    final_dist, max_hold = decide_next_dst(cost, order)
                    if driver_hold < store_hold and current_id > 0 and max_hold == 1: final_dist = 0  
            else: final_dist = 0 #荷物が無いので店に戻る
                
            heading = next_id[current_id][final_dist]
            dist_left = edge[current_id][heading] - 1
            command[t] = heading + 1

    for i in range(T): print(command[i])


    return 0

if __name__ == "__main__":
    solve()