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.*;
import java.io.*;

public class Main {
    public static void main(String args[]) {new Main().run();}

    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(System.out);
    void run(){
        work();
        out.flush();
    }
    long mod=998244353;
    long gcd(long a,long b) {
        return a==0?b:gcd(b%a,a);
    }
    int[][] color;
    ArrayList<int[]> list;
    ArrayList<Integer>[] graph;
    int[] pre;
    int[] dep;
    void work() {
        int n=ni();
        dep=new int[n];
        pre=new int[n];
        list=new ArrayList<>();
        graph=ng(n,n-1);
        color=new int[n][n];
        for(int[] col:color){
            Arrays.fill(col,1);
        }
        dfs(0,new boolean[n],0,-1);
        int m=ni();
        int[][] rec=new int[m][];
        for(int i=0;i<m;i++){
            int n1=ni()-1,n2=ni()-1,k=ni();
            rec[i]=new int[]{n1,n2,k};
            int minDep=Math.min(dep[n1],dep[n2]);
            while(dep[n1]!=minDep){
                int next=pre[n1];
                color[next][n1]=Math.max(k,color[next][n1]);
                color[n1][next]=Math.max(k,color[n1][next]);
                n1=next;
            }
            while(dep[n2]!=minDep){
                int next=pre[n2];
                color[next][n2]=Math.max(k,color[next][n2]);
                color[n2][next]=Math.max(k,color[n2][next]);
                n2=next;
            }
            while(n1!=n2){
                color[pre[n1]][n1]=Math.max(k,color[pre[n1]][n1]);
                color[n1][pre[n1]]=Math.max(k,color[n1][pre[n1]]);
                color[pre[n2]][n2]=Math.max(k,color[pre[n2]][n2]);
                color[n2][pre[n2]]=Math.max(k,color[n2][pre[n2]]);
                n1=pre[n1];
                n2=pre[n2];
            }
        }
        out:
        for(int i=0;i<m;i++){
            int n1=rec[i][0],n2=rec[i][1],k=rec[i][2];
            int minDep=Math.min(dep[n1],dep[n2]);
            while(dep[n1]!=minDep){
                int next=pre[n1];
                if(color[next][n1]==k){
                    continue out;
                }
                n1=next;
            }
            while(dep[n2]!=minDep){
                int next=pre[n2];
                if(color[next][n2]==k){
                    continue out;
                }
                n2=next;
            }
            while(n1!=n2){
                if(color[n1][pre[n1]]==k){
                    continue out;
                }
                if(color[n2][pre[n2]]==k){
                    continue out;
                }
                n1=pre[n1];
                n2=pre[n2];
            }
            out.println(-1);
            return;
        }
        for(int[] l:list){
            out.print(color[l[0]][l[1]]+" ");
        }
    }

    private void dfs(int node, boolean[] vis,int d,int p) {
        dep[node]=d;
        vis[node]=true;
        pre[node]=p;
        for(int nn:graph[node]){
            if(!vis[nn]){
                dfs(nn,vis,d+1,node);
            }
        }
    }

    private ArrayList<long[]>[] ngw(int n, int m) {
        ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
            graph[(int)s].add(new long[] {e,w});
            graph[(int)e].add(new long[] {s,w});
        }
        return graph;
    }

    private ArrayList<Integer>[] ng(int n, int m) {
        ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            int s=in.nextInt()-1,e=in.nextInt()-1;
            list.add(new int[]{s,e});
            graph[s].add(e);
            graph[e].add(s);
        }
        return graph;
    }

    private int ni() {
        return in.nextInt();
    }

    private long nl() {
        return in.nextLong();
    }

    private String ns() {
        return in.next();
    }

    private long[] na(int n) {
        long[] A=new long[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextLong();
        }
        return A;
    }
    private int[] nia(int n) {
        int[] A=new int[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextInt();
        }
        return A;
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;

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


    public String next()
    {
        while(st==null || !st.hasMoreElements())//回车，空行情况
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

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

    public long nextLong()
    {
        return Long.parseLong(next());
    }
}