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 Main {

    static ArrayList<Integer>[] a;
    static int[][] p;
    static int[] f;
    static int ind = 0;
    static int[] ans;
    static TreeMap<component, Integer> tm;
    static int[][] interpreter;


    static void dfs_p(int v, int par) {
        p[ind][v] = par;
        for (int to : a[v]) {
            if (to != par) {
                dfs_p(to, v);
            }
        }
    }

    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner();
        int n = in.nextInt();
        init(n);
        for (int i = 0; i < n - 1; i++) {
            int from = in.nextInt() - 1;
            int to = in.nextInt() - 1;
            a[to].add(from);
            a[from].add(to);
            interpreter[Math.min(to, from)][Math.max(to, from)] = i;
        }
        for (int i = 0; i < n; i++) {
            dfs_p(ind = i, -1);
        }
        int m = in.nextInt();
        point[] q = new point[m];
        for (int i = 0; i < m; i++) {
            q[i] = new point(in.nextInt() - 1, in.nextInt() - 1, in.nextInt());
        }
        Arrays.sort(q, new point());
        for (int i = 0; i < m; i++) {
            int cur = q[i].b;
            while (p[q[i].a][cur] != -1) {
                ans[get(p[q[i].a][cur], cur)] = q[i].val;
                cur = p[q[i].a][cur];
            }
        }
        boolean check = true;
        for (int i = 0; i < m; i++) {
            int cur = q[i].b;
            int min = Integer.MAX_VALUE;
            while (p[q[i].a][cur] != -1) {
                min = Math.min(min, ans[get(p[q[i].a][cur], cur)]);
                cur = p[q[i].a][cur];
            }
            if (min != q[i].val) check = false;
        }
        if (!check) in.out.println(-1);
        else {
            for (int i = 0; i < n - 1; i++) {
                in.out.print(Math.max(ans[i], 1) + " ");
            }
        }
        in.out.close();
    }


    static int get(int x, int y) {
        if (x > y) {
            int t = x;
            x = y;
            y = t;
        }
        return interpreter[x][y];
    }

    static void init(int n) {
        a = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            a[i] = new ArrayList<>();
        }
        tm = new TreeMap<>((o1, o2) -> Integer.compare(o1.x, o2.x) * 2 + Integer.compare(o1.y, o2.y));
        p = new int[n][n];
        f = new int[n];
        ans = new int[n];
        interpreter = new int[n][n];
    }
}


class component {
    int x, y;

    public component(int x, int y) {
        this.x = x;
        this.y = y;
    }
}


class point implements Comparator<point> {
    int a, b, val;

    public point(int a, int b, int val) {
        this.a = a;
        this.b = b;
        this.val = val;
    }

    public point() {
    }

    @Override
    public int compare(point o1, point o2) {
        return Integer.compare(o1.val, o2.val);
    }
}

class FastScanner {
    BufferedReader br;
    StringTokenizer st = new StringTokenizer("");
    PrintWriter out;

    FastScanner() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
    }

    FastScanner(String in, String out_) throws FileNotFoundException {
        br = new BufferedReader(new FileReader(in));
        out = new PrintWriter(out_);
    }

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

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

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