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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct P {
  int v, id;
};
struct PP {
  int a, b, c;
};
const int MAXN = 50 * 100;
vector<P> adj[MAXN];
int vm[MAXN];
P parent[MAXN];
int inter[MAXN][2];
int in = 0;
void parcours(int noeud, int pere, int id) {
  parent[noeud] = {pere, id};
  inter[noeud][0] = in;
  in++;
  for (auto voisin : adj[noeud]) {
    if (voisin.v != pere) {
      parcours(voisin.v, noeud, voisin.id);
    }
  }
  inter[noeud][1] = in;
  in++;
}
bool ancetre(int a, int b) {
  if (inter[a][0] <= inter[b][0] && inter[a][1] >= inter[b][1]) return true;
  return false;
}
void remplir(int actu, int check, int valeur) {
  if (parent[actu].v == -1 || ancetre(actu, check)) {
    return;
  }
  vm[parent[actu].id] = max(vm[parent[actu].id], valeur);
  remplir(parent[actu].v, check, valeur);
}
int minSur(int actu, int check) {
  if (parent[actu].v == -1 || ancetre(actu, check)) {
    return 1e6;
  }
  return min(vm[parent[actu].id], minSur(parent[actu].v, check));
}
int main() {
  int N;
  cin >> N;
  for (int i = 0; i < N - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back({b, i});
    adj[b].push_back({a, i});
  }
  parcours(0, -1, -1);
  int nbQuestions;
  cin >> nbQuestions;
  vector<PP> questions;
  for (int i = 0; i < nbQuestions; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    questions.push_back({a, b, c});
    remplir(a, b, c);
    remplir(b, a, c);
  }
  for (int i = 0; i < N; i++)
    if (vm[i] == 0) vm[i] = 1e6;
  for (int i = 0; i < nbQuestions; i++) {
    int leMin = min(minSur(questions[i].b, questions[i].a),
                    minSur(questions[i].a, questions[i].b));
    if (leMin != questions[i].c) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < N - 1; i++) {
    cout << vm[i] << " ";
  }
  cout << endl;
}