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:
def warshall_floid(graph):
    """warshall-floid法: 各頂点から各頂点への最短距離を求める
    重み付きグラフgraphは隣接リストで与えられる
    計算量: O(|N|^3)
    """
    n = len(graph)
    INF = 10**18
    dp = [[INF]*n for i in range(n)]
    for i in range(n):
        dp[i][i] = 0

    for i in range(n):
        for j, cost in graph[i]:
            dp[i][j] = cost
            dp[j][i] = cost

    for k in range(n):
        for i in range(n):
            for j in range(n):
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j])
    return dp


def trace_route(start, goal):
    """warshall-floid法で求めた最短距離を通るための経路を復元する
    計算量: O(|N|^2)
    """
    pos = start
    ans = []
    while pos != goal:
        ans.append(pos)
        for next_pos in range(n):
            if next_pos != pos and dp[pos][next_pos] + dp[next_pos][goal] == dp[pos][goal]:
                pos = next_pos
                break
    ans.append(goal)
    return ans


# 値を入力する
n, m = map(int, input().split())
info = [list(map(int, input().split())) for i in range(m)]
t_max = int(input())
query = [None] * t_max
for i in range(t_max):
    tmp = int(input())
    query[i] = [list(map(int, input().split())) for i in range(tmp)]

# グラフを構築
graph = [[] for i in range(n)]
for a, b, cost in info:
    a -= 1
    b -= 1
    graph[a].append((b, cost))
    graph[b].append((a, cost))

# 最短距離を求める
dp = warshall_floid(graph)

visited = [False] * n
visited[0] = True
route = [0]
pos = 0
total_cost = 0
while True:
    kyori = 10**9
    next_pos = -1
    for next_p, cost in enumerate(dp[pos]):
        if visited[next_p]:
            continue
        if cost< kyori:
            kyori = cost
            next_pos = next_p
    if next_pos == -1:
        break
    else:
        visited[next_pos] = True
        route.append(next_pos)
        total_cost += kyori

for i in range(n-1):
    pos = route[i]
    next_pos = route[i+1]

for i in range(t_max):
	print(-1)