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 A {

	static int[][][] adj;
	static int[] min;

	static boolean dfs(int u, int p, int target, int value) {
		if (u == target)
			return true;
		for (int[] nxt : adj[u]) {
			int v = nxt[0];
			if (v == p)
				continue;
			if (dfs(v, u, target, value)) {
				min[nxt[1]] = Math.max(min[nxt[1]], value);
				return true;
			}
		}
		return false;

	}

	static pair getMin(int u, int p, int target) {
		if (u == target)
			return new pair();
		pair ans = new pair(Integer.MAX_VALUE, false);
		for (int[] nxt : adj[u]) {
			int v = nxt[0];
			if (v == p)
				continue;
			pair child = getMin(v, u, target);
			if (child.found) {
				ans.found = true;
				ans.min = Math.min(child.min, min[nxt[1]]);
				return ans;
			}
		}
		return ans;

	}

	static class pair {
		int min;
		boolean found;

		pair() {
			min = Integer.MAX_VALUE;
			found = true;
		}

		pair(int x, boolean b) {
			min = x;
			found = b;
		}
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		int n = sc.nextInt();

		adj = new int[n][][];
		min = new int[n];
		int[] deg = new int[n];
		int[] a = new int[n], b = new int[n];
		for (int i = 1; i < n; i++) {
			int u = sc.nextInt() - 1, v = sc.nextInt() - 1;
			a[i] = u;
			b[i] = v;
			deg[u]++;
			deg[v]++;
			min[i] = 1;
		}
		for (int i = 0; i < n; i++)
			adj[i] = new int[deg[i]][2];
		for (int i = 1; i < n; i++) {
			int u = a[i], v = b[i];
			adj[u][--deg[u]] = new int[] { v, i };
			adj[v][--deg[v]] = new int[] { u, i };

		}
		int m = sc.nextInt();
		int[][] queries = new int[m][3];
		for (int i = 0; i < m; i++) {
			int u = sc.nextInt() - 1, v = sc.nextInt() - 1, value = sc.nextInt();
			queries[i] = new int[] { u, v, value };
			dfs(u, u, v, value);
		}
		boolean ok = true;
		for (int[] query : queries) {
			pair check = getMin(query[0], query[0], query[1]);
			if (check.min != query[2]) {
				ok = false;
				break;
			}
		}
		if (ok) {
			for (int i = 1; i < n; i++)
				out.print(min[i] + " ");
		} else
			out.println(-1);
		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

		Scanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

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

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

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

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

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

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

	}

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}

}