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 javax.swing.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class BerlandBeauty {
    ArrayList<int[]>[] adj;
    int[] in;
    int[] out;
    int[][] parentOf;
    int time;
    int[] minBeauty;


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

    public void solve() throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(f.readLine());
        this.adj = new ArrayList[n];
        this.in = new int[n];
        this.out = new int[n];
        this.parentOf = new int[n][2];
        for (int i = 0; i < n; i++) {
            parentOf[i] = new int[2];
        }
        for (int i = 0; i < n; i++) {
            this.adj[i] = new ArrayList<int[]>();
        }
        for (int i = 0; i < n - 1; i++) {
            StringTokenizer tokenizer = new StringTokenizer(f.readLine());
            int a = Integer.parseInt(tokenizer.nextToken()) - 1;
            int b = Integer.parseInt(tokenizer.nextToken()) - 1;
            int[] ar1 = {a, i + 1};
            int[] ar2 = {b, i + 1};
            adj[a].add(ar2);
            adj[b].add(ar1);
        }

        this.time = 0;
        dfs1(0);
        adj = null;

        int m = Integer.parseInt(f.readLine());
        int[][] queries = new int[m][3];
        minBeauty = new int[n];
        Arrays.fill(minBeauty, 1);
        for (int i = 0; i < m; i++) {
            StringTokenizer tokenizer = new StringTokenizer(f.readLine());
            int a = Integer.parseInt(tokenizer.nextToken()) - 1;
            int b = Integer.parseInt(tokenizer.nextToken()) - 1;
            int min = Integer.parseInt(tokenizer.nextToken());
            queries[i][0] = a;
            queries[i][1] = b;
            queries[i][2] = min;
            this.dfs2(a, b, min);
            this.dfs2(b, a, min);
        }

        boolean isPossible = true;
        for (int i = 0; i < m; i++) {
            boolean currentIsGood = checkIfGood(queries[i][0], queries[i][1], queries[i][2]) || checkIfGood(queries[i][1], queries[i][0], queries[i][2]);
            isPossible = isPossible && currentIsGood;
        }

        PrintWriter out = new PrintWriter(System.out);
        if (isPossible) {
            out.print(minBeauty[1]);
            for (int i = 2; i <= n - 1; i ++) {
                out.print(" ");
                out.print(minBeauty[i]);
            }
            out.println();
        } else {
            out.println(-1);
        }

        out.close();
    }

    private void dfs1(int node) {
        this.in[node] = time++;
        for (int[] adjacent : adj[node]) {
            if (adjacent[0] != parentOf[node][0]) {
                parentOf[adjacent[0]][0] = node;
                parentOf[adjacent[0]][1] = adjacent[1];
                dfs1(adjacent[0]);
            }
        }
        this.out[node] = time++;
    }

    private boolean checkIfGood(int startNode, int otherNode, int minimum) {
        while (!(this.in[startNode] <= this.in[otherNode] && out[startNode] >= out[otherNode])) {
            int newNode = parentOf[startNode][0];
            //System.out.println(this.minBeauty[parentOf[startNode][1]] + " " + minimum);
            if (this.minBeauty[parentOf[startNode][1]] == minimum) {
                return true;
            }
            // System.out.println(parentOf[startNode][1] + " " + this.minBeauty[parentOf[startNode][1]]);
            startNode = newNode;
        }
        return false;
    }

    private void dfs2(int startNode, int otherNode, int minimum) {
       // System.out.println("wop " + startNode);
        while (!(this.in[startNode] <= this.in[otherNode] && out[startNode] >= out[otherNode])) {
            int newNode = parentOf[startNode][0];
            this.minBeauty[parentOf[startNode][1]] = Math.max(minBeauty[parentOf[startNode][1]], minimum);
           // System.out.println(parentOf[startNode][1] + " " + this.minBeauty[parentOf[startNode][1]]);
            startNode = newNode;
        }
    }
}