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.*;

@SuppressWarnings("unchecked")
public class Problem_F {
	static List<Pair>[] G;
	static Pair[] par;

	static class Pair {
		int x, y;

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

	static class LCA {
		int N;
		List<Pair>[] G;
		int[][] par;
		int[] dep;

		public LCA(int N, List<Pair>[] G) {
			this.N = N;
			this.G = G;
			par = new int[N + 1][17];
			dep = new int[N + 1];
			for (int i = 1; i <= N; i++) {
				Arrays.fill(par[i], -1);
			}
			dfs(1, -1);
			for (int i = 1; i < 17; i++) {
				for (int j = 1; j <= N; j++) {
					if (par[j][i - 1] == -1) {
						continue;
					}
					par[j][i] = par[par[j][i - 1]][i - 1];
				}
			}
		}

		public void dfs(int x, int p) {
			for (Pair y : G[x]) {
				if (y.x == p) {
					continue;
				}
				par[y.x][0] = x;
				dep[y.x] = dep[x] + 1;
				dfs(y.x, x);
			}
		}

		public int lca(int u, int v) {
			if (dep[u] < dep[v]) {
				int tmp = u;
				u = v;
				v = tmp;
			}
			int diff = dep[u] - dep[v];
			for (int i = 0; i < 17; i++) {
				if (((diff >> i) & 1) == 1) {
					u = par[u][i];
				}
			}
			if (u == v) {
				return u;
			}
			for (int i = 16; i >= 0; i--) {
				if (par[u][i] != -1 && par[u][i] != par[v][i]) {
					u = par[u][i];
					v = par[v][i];
				}
			}
			u = par[u][0];
			return u;
		}
	}

	public static void dfs(int x, int p) {
		for (Pair y : G[x]) {
			if (y.x == p) {
				continue;
			}
			par[y.x] = new Pair(x, y.y);
			dfs(y.x, x);
		}
	}

	public static void main(String[] args) {
		InputReader in = new InputReader();
		StringBuilder out = new StringBuilder();

		int N = in.nextInt();
		G = new List[N + 1];
		for (int i = 1; i <= N; i++) {
			G[i] = new ArrayList<>();
		}
		for (int i = 0; i < N - 1; i++) {
			int x = in.nextInt();
			int y = in.nextInt();
			G[x].add(new Pair(y, i));
			G[y].add(new Pair(x, i));
		}
		int M = in.nextInt();
		int[] a = new int[M];
		int[] b = new int[M];
		int[] g = new int[M];
		for (int i = 0; i < M; i++) {
			a[i] = in.nextInt();
			b[i] = in.nextInt();
			g[i] = in.nextInt();
		}
		LCA lca = new LCA(N, G);
		par = new Pair[N + 1];
		dfs(1, -1);
		int[] f = new int[N - 1];
		Arrays.fill(f, 1);
		for (int i = 0; i < M; i++) {
			int u = lca.lca(a[i], b[i]);
			for (int x : new int[] { a[i], b[i] }) {
				while (u != x) {
					Pair p = par[x];
					f[p.y] = Math.max(f[p.y], g[i]);
					x = p.x;
				}
			}
		}
		boolean flag1 = true;
		for (int i = 0; i < M && flag1; i++) {
			int u = lca.lca(a[i], b[i]);
			boolean flag2 = false;
			for (int x : new int[] { a[i], b[i] }) {
				while (u != x && !flag2) {
					Pair p = par[x];
					if (f[p.y] == g[i]) {
						flag2 = true;
					}
					x = p.x;
				}
			}
			flag1 &= flag2;
		}
		if (!flag1) {
			System.out.println(-1);
			return;
		}
		for (int i = 0; i < N - 1; i++) {
			out.append(f[i]).append(' ');
		}
		out.setCharAt(out.length() - 1, '\n');

		System.out.print(out);
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer st;

		public InputReader() {
			reader = new BufferedReader(new InputStreamReader(System.in));
		}

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

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}

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