Problem:
Score : 700 points

Problem Statement

Squid loves painting vertices in graphs.

There is a simple undirected graph consisting of N vertices numbered 1 through N, and M edges.
Initially, all the vertices are painted in color 0. The i-th edge bidirectionally connects two vertices a_i and b_i. The length of every edge is 1.

Squid performed Q operations on this graph. In the i-th operation, he repaints all the vertices within a distance of d_i from vertex v_i, in color c_i.

Find the color of each vertex after the Q operations.

Constraints

1 ≤ N,M,Q ≤ 10^5

1 ≤ a_i,b_i,v_i ≤ N

a_i ≠ b_i

0 ≤ d_i ≤ 10

1 ≤ c_i ≤10^5

d_i and c_i are all integers.

There are no self-loops or multiple edges in the given graph.

Partial Score

200 points will be awarded for passing the testset satisfying 1 ≤ N,M,Q ≤ 2{,}000.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
:
a_{M} b_{M}
Q
v_1 d_1 c_1
:
v_{Q} d_{Q} c_{Q}

Output

Print the answer in N lines.
In the i-th line, print the color of vertex i after the Q operations.

Sample Input 1

7 7
1 2
1 3
1 4
4 5
5 6
5 7
2 3
2
6 1 1
1 2 2

Sample Output 1

2
2
2
2
2
1
0

Initially, each vertex is painted in color 0.
In the first operation, vertices 5 and 6 are repainted in color 1.
In the second operation, vertices 1, 2, 3, 4 and 5 are repainted in color 2.

Sample Input 2

14 10
1 4
5 7
7 11
4 10
14 7
14 3
6 14
8 11
5 13
8 3
8
8 6 2
9 7 85
6 9 3
6 7 5
10 3 1
12 9 4
9 6 6
8 2 3

Sample Output 2

1
0
3
1
5
5
3
3
6
1
3
4
5
3

The given graph may not be connected.

Solution in Python (2.7.6) (judge verdict: Time Limit Exceeded)

from collections import defaultdict,  deque

visited = [False]*(10**6 + 1)
color = [0]*(10**6 + 1)
level = [None]*(10**6 + 1)

class Graph:
  
    def __init__(self,vertices):
        self.V= vertices
        self.graph = defaultdict(list)
  
    def addEdge(self,u,v):
        self.graph[u].append(v)
        self.graph[v].append(u)
        
    def rmvEdge(self, u, v):
        for index, key in enumerate(self.graph[u]):
            if key == v:
                self.graph[u].pop(index)
        for index, key in enumerate(self.graph[v]):
            if key == u:
                self.graph[v].pop(index)
                
def BFS(graph, start, colorTemp, depth):
    queue = deque()
    queue.append(start)
    color[start] = colorTemp
    visited[start] = True
    level[start] = 0
    while queue:
        s = queue.popleft()
        if level[s] == depth:
            break
        for i in graph[s]:
            if not visited[i]:
                level[i] = level[s] + 1
                queue.append(i)
                color[i] = colorTemp
                visited[i] = True
    
n, m = map(int, raw_input().split())

g = Graph(n)

for _ in xrange(m):
    a, b = map(int, raw_input().split())
    g.addEdge(a, b)

q = int(raw_input())

for _ in xrange(q):
    v, d, c = map(int, raw_input().split())
    visited = [False]*(10**6 + 1)
    level = [None]*(10**6 + 1)
    BFS(g.graph, v, c, d)
    #print color[1: n+1]

for i in xrange(1, n+1):
    print color[i]