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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

/***********************
*  Date: 2018.11.28    *
*  Problem:            *
*  Author: xuziyuan    * 
***********************/

#include <bits/stdc++.h>

#define rt0 return 0
#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define replet(c) for(char c='a';c<='z';c++)
#define LL long long
#define pii pair <int,int>
#define pb push_back
#define fi first
#define se second
#define mpr make_pair
#define sqr(a) ((a)*(a))

using namespace std;

const LL MOD=1e9+7;

int n,m,q,appe[100010][20],coap[100010];//appe:appearance
vector <int> g[100010];

int main()
{
	cin>>n>>m;
	rep(i,m)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		g[x].pb(y);
		g[y].pb(x);
	}
	cin>>q;
	repn(i,q)
	{
		int x,y;
		scanf("%d%d%d",&x,&y,&coap[i]);
		appe[x][y]=i;
	}
	for(int i=9;i>=0;i--)//算能不能覆盖appe[j][i]之前要先算出appe[p][i+1]会不会被覆盖，所以i要倒着循环 
	{
		repn(p,n)
		{
			rep(k,g[p].size())//i在这里代表可以控制的范围
			{
				int j=g[p][k];
				appe[j][i]=max(appe[j][i],appe[p][i+1]);/*远一条边距离就要加1，这里是看后者可不可以覆盖前者，
				可以说是一种dp，一个类似递归的过程，如果再之前的点把后者覆盖了，也会在前者上体现出来*/
			}	
		}	
	} 
	repn(i,n)
	{
		int lst=0;
		rep(j,11) lst=max(lst,appe[i][j]);
		cout<<coap[lst]<<endl;
	}
	rt0;
}