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;
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream& operator<<(ostream& os, set<T> second) {
  os << "{ ";
  for (auto s : second) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream& operator<<(ostream& os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " |";
  __f(comma + 1, args...);
}
const int N = 5000;
vector<bool> vis(N);
vector<pair<int, int>> adj[N];
vector<int> wgt(N);
void dfs(int node) {
  vis[node] = true;
  for (pair<int, int> childPr : adj[node]) {
    if (!vis[childPr.first]) dfs(childPr.first);
  }
}
int getMinWt(int u, int v, int par = -1) {
  for (pair<int, int> childPr : adj[u]) {
    if (childPr.first == par)
      continue;
    else if (childPr.first == v)
      return wgt[childPr.second];
    int wt = getMinWt(childPr.first, v, u);
    if (wt >= 0) return min(wt, wgt[childPr.second]);
  }
  return -1;
}
void solve() {
  int n;
  cin >> n;
  vector<pair<int, int>> egs(n - 1);
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    egs[i] = {x, y};
    adj[x].push_back({y, i});
    adj[y].push_back({x, i});
  }
  int m;
  cin >> m;
  vector<pair<pair<int, int>, int>> res(m);
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    res[i] = {{u, v}, w};
  }
  for (int i = 0; i < n - 1; ++i) {
    auto eg = egs[i];
    fill(vis.begin(), vis.begin() + n, false);
    vis[eg.second] = true;
    dfs(eg.first);
    vis[eg.second] = false;
    for (auto resEgPr : res) {
      pair<int, int>& resEg = resEgPr.first;
      if (vis[resEg.first] != vis[resEg.second])
        wgt[i] = max(wgt[i], resEgPr.second);
    }
    if (wgt[i] == 0) wgt[i] = 1e6;
  }
  for (auto resEgPr : res) {
    pair<int, int>& resEg = resEgPr.first;
    int mnWt = getMinWt(resEg.first, resEg.second);
    if (mnWt != resEgPr.second) {
      cout << "-1\n";
      return;
    }
  }
  for (int i = 0; i < n - 1; ++i) {
    cout << wgt[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1, tc = 1;
  while (T--) {
    solve();
  }
  return 0;
}