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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char* x) { cerr << '\"' << x << '\"'; }
void __print(const string& x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V>& x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T& x) {
  int f = 0;
  cerr << '{';
  for (auto& i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int M = 1e9 + 7;
vector<vector<pair<int, int>>> adj;
vector<int> eval;
bool dfs(int i, int p, int en, int val) {
  if (i == en) {
    return true;
  }
  for (auto j : adj[i]) {
    if (j.first == p) {
      continue;
    }
    if (dfs(j.first, i, en, val)) {
      eval[j.second] = max(eval[j.second], val);
      return true;
    }
  }
  return false;
}
int cdfs(int i, int p, int en, int cur) {
  if (i == en) {
    return cur;
  }
  for (auto j : adj[i]) {
    if (j.first == p) {
      continue;
    }
    int tcur = min(cur, eval[j.second]);
    int temp = cdfs(j.first, i, en, tcur);
    if (temp != 0) {
      return temp;
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  adj = vector<vector<pair<int, int>>>(n);
  eval = vector<int>(n - 1, -1);
  int u, v;
  for (int i = 0; i < n - 1; ++i) {
    cin >> u >> v;
    --u;
    --v;
    adj[u].push_back({v, i});
    adj[v].push_back({u, i});
  }
  int m;
  cin >> m;
  int val;
  vector<array<int, 3>> B(m);
  for (int i = 0; i < m; ++i) {
    cin >> u >> v >> val;
    --u;
    --v;
    B[i] = {u, v, val};
    dfs(u, -1, v, val);
  }
  for (int i = 0; i < n - 1; ++i) {
    if (eval[i] == -1) {
      eval[i] = 1e9;
    }
  }
  for (int i = 0; i < m; ++i) {
    if (cdfs(B[i][0], -1, B[i][1], 1e9) != B[i][2]) {
      cout << -1 << "\n";
      return 0;
    }
  }
  for (int i = 0; i < n - 1; ++i) {
    if (eval[i] == 1e9) {
      eval[i] = 1e6;
    }
    cout << eval[i] << " ";
  }
  cout << "\n";
  return 0;
}