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>
#pragma comment(linker, "/STACK:268435456")
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2>& p) {
  out << p.first << " " << p.second;
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& v) {
  for (int i = 0; i < int(v.size()); ++i) {
    in >> v[i];
  }
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T>& v) {
  for (int i = 0; i < int(v.size()); ++i) {
    out << v[i] << ' ';
  }
  return out;
}
struct Node {
  int was = -1;
  vector<int> to;
};
vector<Node> g;
struct Edge {
  int u;
  int v;
  int f = 1;
};
vector<Edge> e;
struct Request {
  int from;
  int to;
  int mf;
  vector<int> path;
  friend const bool operator<(const Request& a, const Request& b) {
    return a.mf < b.mf;
  }
  friend istream& operator>>(istream& in, Request& r) {
    in >> r.from >> r.to >> r.mf;
    --r.from;
    --r.to;
    return in;
  }
};
vector<Request> r;
bool find_path(int v, int to, int rn) {
  if (g[v].was == rn) return false;
  g[v].was = rn;
  if (v == to) return true;
  for (int i = 0; i < int(g[v].to.size()); ++i) {
    if (find_path(e[g[v].to[i]].u, to, rn)) {
      r[rn].path.push_back(g[v].to[i]);
      return true;
    }
    if (find_path(e[g[v].to[i]].v, to, rn)) {
      r[rn].path.push_back(g[v].to[i]);
      return true;
    }
  }
  return false;
}
void fail() {
  cout << -1 << endl;
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n;
  g.resize(n);
  e.resize(n - 1);
  for (int i = 0; i < int(n - 1); ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    e[i].u = a;
    e[i].v = b;
    g[a].to.push_back(i);
    g[b].to.push_back(i);
  }
  cin >> m;
  r.resize(m);
  cin >> r;
  sort((r).begin(), (r).end());
  for (int i = 0; i < int(m); ++i)
    if (!find_path(r[i].from, r[i].to, i)) fail();
  for (int i = 0; i < int(m); ++i) {
    for (int j = 0; j < int(r[i].path.size()); ++j) e[r[i].path[j]].f = r[i].mf;
  }
  for (int i = 0; i < int(m); ++i) {
    int mf = 1e9;
    for (int j = 0; j < int(r[i].path.size()); ++j)
      mf = min(mf, e[r[i].path[j]].f);
    if (mf != r[i].mf) fail();
  }
  for (int i = 0; i < int(n - 1); ++i) cout << e[i].f << ' ';
  return 0;
}