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.lang.reflect.Array;
import java.util.*;

import static java.lang.Math.*;

public class Main {
    ArrayList<Integer> graph[];
    int[][] p;
    int put;

    void dfs(int v, int parent) {
        p[put][v] = parent;
        for (int to : graph[v]) {
            if (to != parent) dfs(to, v);
        }
    }

    void run() throws IOException {
        int n = nextInt();
        graph = new ArrayList[n];
        p = new int[n][n];
        for (int i = 0; i < graph.length; i++) {
            graph[i] = new ArrayList<>();
        }
        int[][] num = new int[n][n];
        for (int i = 0; i < n - 1; i++) {
            int from = nextInt() - 1;
            int to = nextInt() - 1;
            graph[from].add(to);
            graph[to].add(from);
            num[from][to] = num[to][from] = i;
        }
        for (int i = 0; i < n; i++) {
        }
        HashSet<Integer> ts = new HashSet<>();
        int q = nextInt();
        int[] f = new int[n - 1];
        int[][] zap = new int[q][3];
        for (int i = 0; i < q; i++) {
            int from = nextInt() - 1;
            int to = nextInt() - 1;
            if (!ts.contains(from) && ts.contains(to)) {
                int x = from;
                from = to;
                to = x;
            }
            if (!ts.contains(from) && !ts.contains(to)) {
                ts.add(from);
                put = from;
                dfs(from, -1);
            }
            int min = nextInt();
            zap[i][0] = from;
            zap[i][1] = to;
            zap[i][2] = min;
            while (p[from][to] != -1) {
                int x = num[p[from][to]][to];
                f[x] = max(f[x], min);
                to = p[from][to];
            }
        }
        ts = new HashSet<>();
        for (int i = 0; i < zap.length; i++) {
            int from = zap[i][0];
            int to = zap[i][1];
            int min = 1000000;
            while (p[from][to] != -1) {
                int x = num[p[from][to]][to];
                min = min(min, f[x]);
                to = p[from][to];
            }
            if (min != zap[i][2]) {
                pw.println(-1);
                pw.close();
                return;
            }
        }
        for (int i = 0; i < f.length; i++) {
            pw.print(max(f[i], 1) + " ");
        }
        pw.close();
    }

    class pair implements Comparable<pair> {
        int from, to;

        public pair(int from, int to) {
            this.from = from;
            this.to = to;
        }

        @Override
        public int compareTo(pair o) {
            if (o.from == this.from) return -Integer.compare(o.to, this.to);
            return -Integer.compare(o.from, this.from);
        }
    }


    Scanner sc = new Scanner(System.in);
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    //BufferedReader br = new BufferedReader(new FileReader("downloading.in"));

    StringTokenizer st = new StringTokenizer("");
    PrintWriter pw = new PrintWriter(System.out);
    //PrintWriter pw = new PrintWriter("downloading.out");

    int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    String next() throws IOException {
        if (!st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    public Main() throws FileNotFoundException {
    }

    public static void main(String[] args) throws IOException {
        new Main().run();
    }
}