Problem: There are n railway stations in Berland. They are connected to each other by n-1 railway sections. The railway network is connected, i.e. can be represented as an undirected tree.

You have a map of that network, so for each railway section you know which stations it connects.

Each of the n-1 sections has some integer value of the scenery beauty. However, these values are not marked on the map and you don't know them. All these values are from 1 to 10^6 inclusive.

You asked m passengers some questions: the j-th one told you three values:

  * his departure station a_j; 
  * his arrival station b_j; 
  * minimum scenery beauty along the path from a_j to b_j (the train is moving along the shortest path from a_j to b_j). 



You are planning to update the map and set some value f_i on each railway section — the scenery beauty. The passengers' answers should be consistent with these values.

Print any valid set of values f_1, f_2, ..., f_{n-1}, which the passengers' answer is consistent with or report that it doesn't exist.

Input

The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of railway stations in Berland.

The next n-1 lines contain descriptions of the railway sections: the i-th section description is two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i), where x_i and y_i are the indices of the stations which are connected by the i-th railway section. All the railway sections are bidirected. Each station can be reached from any other station by the railway.

The next line contains a single integer m (1 ≤ m ≤ 5000) — the number of passengers which were asked questions. Then m lines follow, the j-th line contains three integers a_j, b_j and g_j (1 ≤ a_j, b_j ≤ n; a_j ≠ b_j; 1 ≤ g_j ≤ 10^6) — the departure station, the arrival station and the minimum scenery beauty along his path.

Output

If there is no answer then print a single integer -1.

Otherwise, print n-1 integers f_1, f_2, ..., f_{n-1} (1 ≤ f_i ≤ 10^6), where f_i is some valid scenery beauty along the i-th railway section.

If there are multiple answers, you can print any of them.

Examples

Input


4
1 2
3 2
3 4
2
1 2 5
1 3 3


Output


5 3 5


Input


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


Output


5 3 1 2 1 


Input


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


Output


-1

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.StringTokenizer;

public class berland {
	static int n;
	static ArrayDeque<Edge>[] adj;
	
	static int[] depth;
	static int[] edgew;
	static Edge[] parEdge;
	static void root(int cur, int par) {
		if(cur!=par)depth[cur]=depth[par]+1;
		for(Edge e:adj[cur]) {
			if(e.v!=par) {
				root(e.v,cur);
				parEdge[e.v]=new Edge(cur,e.id);
			}
		}
	}
	
	public static void main(String[] args) {
		FS s = new FS();
		int n = s.nextInt();
		Arrays.setAll(adj=new ArrayDeque[n],i->new ArrayDeque<>());
		int gidx=0;
		for(int i=0;i<n-1;i++) {
			int u=s.nextInt()-1;
			int v=s.nextInt()-1;
			Edge e=new Edge(v,gidx++);
			adj[u].add(e);
			adj[v].add(new Edge(u,e.id));
		}
		depth=new int[n];
		parEdge=new Edge[n];
		root(0,0);
//		System.out.println("parsing queries");
		int q=s.nextInt();
		int[][] queries=new int[q][3];
		Arrays.setAll(queries,i->new int[] {s.nextInt()-1,s.nextInt()-1,s.nextInt()});
		Arrays.sort(queries,(a,b)->a[2]-b[2]);
		Arrays.fill(edgew=new int[n-1],-1);
//		System.out.println("query time");
		
		for(int[] qr:queries) {
			
			int u=qr[0],v=qr[1],w=qr[2];
			if(depth[v]>depth[u]) {
				int temp=v;
				v=u;
				u=temp;
			}
			
			//u is guaranteed deeper or equal
			while(depth[u]!=depth[v]) {
				edgew[parEdge[u].id]=w;
				u=parEdge[u].v;
//				System.out.println(u+","+depth[u]+", and "+v+","+depth[v]);
			}
//			System.out.println(u+","+v+","+w);
			while(u!=v) {
				edgew[parEdge[u].id]=w;
				u=parEdge[u].v;
				edgew[parEdge[v].id]=w;
				v=parEdge[v].v;
			}
			
		}
		
//		System.out.println("checking now");
		for(int[] qr:queries) {
			int u=qr[0],v=qr[1],w=qr[2];
			if(depth[v]>depth[u]) {
				int temp=v;
				v=u;
				u=temp;
			}
			int min=Integer.MAX_VALUE;
			//u is guaranteed deeper or equal
			while(depth[u]!=depth[v]) {
				min=Math.min(edgew[parEdge[u].id],min);
				u=parEdge[u].v;
			}
			while(u!=v) {
				min=Math.min(edgew[parEdge[u].id],min);
				u=parEdge[u].v;
				min=Math.min(edgew[parEdge[v].id],min);
				v=parEdge[v].v;
			}
			if(min!=w) {
				System.out.println(-1);
				return;
			}
			
		}
		for(int x:edgew)System.out.print(x==-1?69+" ":x+" ");
		
	}

	static class Edge{
		int v,id;
		Edge(int u, int id){
			v=u;
			this.id=id;
		}
	}
	
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		
		String next() {
			while (!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (Exception e) {
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}