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.util.*;
import java.io.*;
import static java.lang.System.*;

public class Hell {

    static final int N = 5005;
    static int n, m, u, v, st, en, cs;
    static ArrayList<ArrayList<pair>> edges = new ArrayList<ArrayList<pair>>();
    static int arr[] = new int[N];
    static int a[] = new int[N];
    static int b[] = new int[N];
    static int c[] = new int[N];

    static class pair {

        private int l, r;

        pair(int l, int r) {
            this.l = l;
            this.r = r;
        }

        int first() {
            return l;
        }

        int second() {
            return r;
        }
    }

    static boolean dfs(int node, int pre) {
        if (node == en) {
            return true;
        }
        for (int i = 0; i < edges.get(node).size(); i++) {
            pair to = edges.get(node).get(i);
            if (to.first() != pre && dfs(to.first(), node)) {
                arr[to.second()] = Math.max(arr[to.second()], cs);
                return true;
            }
        }
        return false;
    }

    static int Find(int node, int pre) {
        if (node == en) {
            return Integer.MAX_VALUE;
        }
        for (int i = 0; i < edges.get(node).size(); i++) {
            pair to = edges.get(node).get(i);
            if (to.first() != pre) {
                int Save = Find(to.first(), node);
                if (Save != Integer.MIN_VALUE) {
                    return Math.min(arr[to.second()], Save);
                }
            }
        }
        return Integer.MIN_VALUE;
    }

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        n = sc.nextInt();
        for (int i = 0; i < N; i++) {
            edges.add(new ArrayList<pair>());
        }
        for (int i = 0; i < n - 1; i++) {
            u = sc.nextInt();
            v = sc.nextInt();
            arr[i] = 1;
            edges.get(u).add(new pair(v, i));
            edges.get(v).add(new pair(u, i));
        }
        m = sc.nextInt();
        for (int i = 0; i < m; i++) {
            a[i] = st = sc.nextInt();
            b[i] = en = sc.nextInt();
            c[i] = cs = sc.nextInt();
            dfs(st, -1);
        }
        for (int i = 0; i < m; i++) {
            st = a[i];
            en = b[i];
            cs = c[i];
            if (Find(st, -1) != cs) {
                System.out.print("-1\n");
                exit(0);
            }
        }
        for (int i = 0; i < n - 1; i++) {
            System.out.printf("%d ", arr[i]);
        }
        System.out.print("\n");
    }

    static class Scanner {

        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

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

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

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

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}