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: Time Limit Exceeded)

#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define pb push_back
#define pp pop_back
#define mp make_pair
#define fi first
#define se second
#define sz(_a) _a.size()
#define all(_a) _a.begin(),_a.end()
#define forn(_i,_j,_k) for(int _i=_j;_i<_k;_i++)
#define pi 3.14159265
#define vi vector<int>
#define vii vector<vi >
#define pii pair<int,int>
#define vpii vector<pii >
using namespace std;
int n,m,col,t[100005]={0},di;
vi v[100005],ve,d,co;
int visit[100005]={0},ans[100005]={0};
void dfs(int pos,int dis,int la){
	if(dis>di)return ;
	if(visit[pos]==0){
		visit[pos]=1;
		ans[pos]=col;
	}
	forn(i,0,sz(v[pos]))if(v[pos][i]!=la)dfs(v[pos][i],dis+1,pos);
}
bool check(){
	forn(i,1,n+1)if(visit[i]==0)return 0;
	return 1;
}
void print(){
	forn(i,1,n+1)cout<<ans[i]<<endl;
}
int main(void){
	forn(i,0,100005)t[i]--;
	cin>>n>>m;
	forn(i,0,m){
		int a,b;
		cin>>a>>b;
		v[a].pb(b);
		v[b].pb(a);
	} 
	int q;cin>>q;
	forn(i,0,q){
		int a,b,c;cin>>a>>b>>c;
		ve.pb(a);d.pb(b);co.pb(c);
	}
	reverse(all(ve));reverse(all(d));reverse(all(co));
	forn(i,0,q){
		if(d[i]<=t[ve[i]])continue;
		t[ve[i]]=d[i];
		col=co[i];di=d[i];
		dfs(ve[i],0,0);
		if(i%10==0&&check())break;
	}
	print();
	return 0;
}