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.

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.

Examples

Input

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


Output

2
2
2
2
2
1
0


Input

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


Output

1
0
3
1
5
5
3
3
6
1
3
4
5
3
from sys import stdin, stderr
from collections import deque

def solve():
    N, M = map(int, input().split())

    Adj = [[] for i in range(N)]

    for i in range(M):
        a, b = map(int, stdin.readline().split())
        a, b = a - 1, b - 1
        Adj[a].append(b)
        Adj[b].append(a)

    Q = int(stdin.readline())

    v = [0] * Q
    d = [0] * Q
    c = [0] * Q

    for i in range(Q):
        v[i], d[i], c[i] = map(int, stdin.readline().split())
        v[i] -= 1

    v.reverse()
    d.reverse()
    c.reverse()

    cols = [0] * N
    nuta = [-1] * N

    for i in range(Q):
        bfs(N, Adj, v[i], d[i], c[i], cols, nuta)

    print(*cols, sep='\n')

def bfs(N, Adj, v, d, c, cols, nuta):
    if nuta[v] >= d:
        return

    nxts = deque([[v, d]])
    nuta[v] = d

    while nxts:
        u, step = nxts.popleft()

        if step < 0:
            continue

        if cols[u] == 0:
            cols[u] = c

        for w in Adj[u]:
            if nuta[w] < step - 1:
                nuta[w] = step - 1
                nxts.append([w, step - 1])

    return

if __name__ == '__main__':
    solve()