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.*;
import java.util.*;

public class f{

    static int N, M, d[][], be[][], u[], v[];
    static ArrayList<Integer> graph[];
    static q qs[];

    static void dfs(int s, int id, int p, int di){
        d[s][id] = p;
        for(int i : graph[id]){
            if(i == p) continue;
            dfs(s, i, id, di+1);
        }
    }

    public static void main(String args[]) throws IOException{
        BufferedReader eat = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter moo = new PrintWriter(System.out);
        N = Integer.parseInt(eat.readLine());
        graph = new ArrayList[N];
        for(int i = 0; i < N; i++) graph[i] = new ArrayList<Integer>();
        u = new int[N-1];
        v = new int[N-1];
        for(int i = 0; i < N-1; i++){
            StringTokenizer st = new StringTokenizer(eat.readLine());
            int x = Integer.parseInt(st.nextToken()) - 1, y = Integer.parseInt(st.nextToken()) - 1;
            u[i] = x;
            v[i] = y;
            graph[x].add(y);
            graph[y].add(x);
        }
        d = new int[N][N];
        for(int i = 0; i < N; i++){
            dfs(i, i, -1, 0);
        }
        M = Integer.parseInt(eat.readLine());
        qs = new q[M];
        for(int i = 0; i < M; i++){
            StringTokenizer st = new StringTokenizer(eat.readLine());
            int a = Integer.parseInt(st.nextToken()) - 1, b = Integer.parseInt(st.nextToken()) - 1, g = Integer.parseInt(st.nextToken());
            qs[i] = new q(a, b, g);
        }
        Arrays.sort(qs);
        be = new int[N][N];
        boolean f = false;
        for(int i = 0; i < M; i++){
            int a = qs[i].a, b = qs[i].b, g = qs[i].g;
            f = false;
            while(a != b){
                int nxt = d[b][a];
                if(be[a][nxt] <= g){
                    f = true;
                    be[a][nxt] = be[nxt][a] = g;
                }
                a = nxt;
            }
            if(!f){
                break;
            }
        }
        if(f){
            for(int i = 0; i < N-1; i++){
                int x = be[u[i]][v[i]];
                moo.print((x == 0 ? 1 : x) + " ");
            }
            moo.println();
        }else{
            moo.println(-1);
        }
        eat.close(); moo.close();
    }

    static class q implements Comparable<q>{
        int a, b, g;
        q(int a, int b, int g){
            this.a = a;
            this.b = b;
            this.g = g;
        }
        public int compareTo(q x){
            return x.g - this.g;
        }
    }
}