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.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Scanner;


public class Main
{

    static int mod = ( int ) 1e9 + 7;
    static ArrayList<Edge>[] adj;
    static int[][] p;

    public static void main( String[] args )
    {
        Scanner sc = new Scanner( System.in );
        int n = sc.nextInt();
        edges = new LinkedList[ n ];
        for( int i = 0; i < n; i++ )
        {
            edges[ i ] = new LinkedList<>();
        }
        for( int i = 1; i < n; i++ )
        {
            int a = sc.nextInt() - 1;
            int b = sc.nextInt() - 1;
            edges[ a ].add( new Edge( b, i - 1 ) );
            edges[ b ].add( new Edge( a, i - 1 ) );
        }
        par = new int[ n ];
        depth = new int[ n ];
        edgeid = new int[ n ];
        Arrays.fill( edgeid, -1 );
        ret = new int[ n - 1 ];
        dfs( 0, -1 );
        Arrays.fill( ret, 1 );
        int m = sc.nextInt();
        int[] lhs = new int[ m ];
        int[] rhs = new int[ m ];
        int[] w = new int[ m ];
        for( int i = 0; i < m; i++ )
        {
            lhs[ i ] = sc.nextInt() - 1;
            rhs[ i ] = sc.nextInt() - 1;
            w[ i ] = sc.nextInt();
            int a = lhs[ i ];
            int b = rhs[ i ];
            while( a != b )
            {
                if( depth[ a ] > depth[ b ] )
                {
                    ret[ edgeid[ a ] ] = Math.max( ret[ edgeid[ a ] ], w[ i ] );
                    a = par[ a ];
                }
                else if( depth[ b ] > depth[ a ] )
                {
                    ret[ edgeid[ b ] ] = Math.max( ret[ edgeid[ b ] ], w[ i ] );
                    b = par[ b ];
                }
                else
                {
                    ret[ edgeid[ a ] ] = Math.max( ret[ edgeid[ a ] ], w[ i ] );
                    a = par[ a ];
                    ret[ edgeid[ b ] ] = Math.max( ret[ edgeid[ b ] ], w[ i ] );
                    b = par[ b ];
                }
            }
        }
        boolean good = true;
        for( int i = 0; good && i < m; i++ )
        {
            int minw = ( int ) 1e6;
            int a = lhs[ i ];
            int b = rhs[ i ];
            while( a != b )
            {
                if( depth[ a ] > depth[ b ] )
                {
                    minw = Math.min( minw, ret[ edgeid[ a ] ] );
                    a = par[ a ];
                }
                else if( depth[ b ] > depth[ a ] )
                {
                    minw = Math.min( minw, ret[ edgeid[ b ] ] );
                    b = par[ b ];
                }
                else
                {
                    minw = Math.min( minw, ret[ edgeid[ a ] ] );
                    a = par[ a ];
                    minw = Math.min( minw, ret[ edgeid[ b ] ] );
                    b = par[ b ];
                }
            }
            good = minw == w[ i ];
        }
        if( !good )
        {
            System.out.println( -1 );
        }
        else
        {
            StringBuilder sb = new StringBuilder();
            for( int i = 0; i < ret.length; i++ )
            {
                sb.append( ret[ i ] ).append(" ");
            }
            System.out.println( sb );
        }
    }
    static LinkedList<Edge>[] edges;
    static int[] ret;
    static int[] par;
    static int[] depth;
    static int[] edgeid;

    public static void dfs( int curr, int p )
    {
        edges[ curr ].stream().filter( ( out ) -> !( out.d == p ) ).map( ( out ) ->
        {
            par[ out.d ] = curr;
            return out;
        } ).map( ( out ) ->
        {
            depth[ out.d ] = 1 + depth[ curr ];
            return out;
        } ).map( ( out ) ->
        {
            edgeid[ out.d ] = out.i;
            return out;
        } ).forEachOrdered( ( out ) ->
        {
            dfs( out.d, curr );
        } );
    }
}

class Edge
{

    public int d, i;

    public Edge( int a, int b )
    {
        d = a;
        i = b;
    }
}