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.Scanner;
import java.util.*;

 
public class F {
    static ArrayList< ArrayList<Integer> > g;
    static int[] parent;
    static int[] h;
    static int[][] a;
    static int glo;

    public static void dfs(int v, int from) {
        //System.out.println("v=" + S(v) + " from=" + S(from));
        parent[v] = from;
        h[v] = h[from] + 1;
        for(int i = 0; i < g.get(v).size(); ++i) {
            int w = g.get(v).get(i);
            if (w == from) continue;

            dfs(w, v);
        }
    }

    static int update(int v, int c) {
        //System.out.println("Update: v=" + S(v) + " color=" + S(c));

        int p = parent[v];
        a[v][p] = c;
        a[p][v] = c;
        return p;
    }

    static int check(int v, int c) {
        //System.out.println("Check: v=" + S(v) + " color=" + S(c));

        int p = parent[v];

        glo = Math.min(glo, a[v][p]);

        return p;
    }


    static String S(int x) {
        return (new Integer(x)).toString();
    }

	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        ArrayList<Edge> e = new ArrayList<Edge>();
        ArrayList<Edge> original = new ArrayList<Edge>();

        int[] ex = new int[n - 1];
        int[] ey = new int[n - 1];
        parent = new int[n + 1];

        a = new int[n + 1][];
        for(int i = 0; i <= n; ++i) a[i] = new int[n + 1];

        h = new int[n + 1];

        g = new ArrayList< ArrayList<Integer> >();
        for(int i = 0; i <= n; ++i) {
            g.add( new ArrayList<Integer>() );
        }


        for(int i = 0; i < n - 1; ++i) {
            ex[i] = sc.nextInt();
            ey[i] = sc.nextInt();
            g.get(ex[i]).add( new Integer(ey[i]) );
            g.get(ey[i]).add( new Integer(ex[i]) );
        }

        int m = sc.nextInt();

        for(int i = 0; i < m; ++i) {
            
            Edge t = new Edge();

            t.x = sc.nextInt();
            t.y = sc.nextInt();
            t.c = sc.nextInt();

            e.add(t);
            original.add( t );            
        }

        Collections.sort(e);

        dfs(1, 1);

        for(int i = 0; i < m; ++i) {
            int x = e.get(i).x;
            int y = e.get(i).y;
            int c = e.get(i).c;

            //System.out.println("x=" + S(x) + " y=" + S(y) + " c=" + S(c));

            if (h[x] < h[y]) {
                int tmp = x;
                x = y;
                y = tmp;
            }

            while (h[x] != h[y]) {
                x = update(x, c);
            }

            while (x != y) {
                x = update(x, c);
                y = update(y, c);
            }
        }

        Boolean ok = true;


        for(int i = 0; i < m; ++i) {
            int x = e.get(i).x;
            int y = e.get(i).y;
            int c = e.get(i).c;

            //System.out.println("x=" + S(x) + " y=" + S(y) + " c=" + S(c));

            if (h[x] < h[y]) {
                int tmp = x;
                x = y;
                y = tmp;
            }
            glo = 1000000000;

            while (h[x] != h[y]) {
                x = check(x, c);
            }

            while (x != y) {
                x = check(x, c);
                y = check(y, c);
            }

            if (glo != c) ok = false;
        }



        if (ok) {
            for(int i = 0; i < n - 1; ++i) {
                int t = a[ex[i]][ey[i]];
                if (t == 0) t++;
                System.out.print(t);
                System.out.print(" ");
            }
            System.out.println("");
        } else {
            System.out.println(-1);
        }

        	
		sc.close();
	}
 
}

    class Edge implements Comparable<Edge> {
        int x;
        int y;
        int c;

        Edge() {
            x = 0;
            y = 0;
            c = 0;
        }

        @Override
        public int compareTo(Edge other) {
            if (c < other.c) return -1;
            if (c > other.c) return 1;
            return 0;
        }

        @Override
        public String toString() {
            return "ok";
        }
    }