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;
const int MAX_V = 5002;
vector<int> G[MAX_V];
int Gm[MAX_V][MAX_V];
int mini = 1e9 + 7;
bool fill_dfs(int v, int t, int w, int prev = -1) {
  if (v == t) return true;
  for (auto e : G[v]) {
    if (e == prev) continue;
    if (fill_dfs(e, t, w, v)) {
      Gm[e][v] = w, Gm[v][e] = w;
      return true;
    }
  }
  return false;
}
bool query_dfs(int v, int t, int prev = -1) {
  if (v == t) return true;
  for (auto e : G[v]) {
    if (e == prev) continue;
    if (query_dfs(e, t, v)) {
      mini = min(Gm[e][v], mini);
      return true;
    }
  }
  return false;
}
int main() {
  int n;
  cin >> n;
  n--;
  vector<pair<int, int>> input(n);
  for (int i = 0; i < n; i++) {
    int a, b;
    cin >> a >> b;
    input[i] = {a, b};
    G[a].push_back(b);
    G[b].push_back(a);
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> g;
  for (int i = 0; i < m; i++) {
    int a, b, v;
    cin >> a >> b >> v;
    g.push_back({v, {a, b}});
  }
  sort(g.begin(), g.end());
  for (auto e : g) fill_dfs(e.second.first, e.second.second, e.first);
  bool ans = true;
  for (auto e : g) {
    mini = 1e9 + 7;
    query_dfs(e.second.first, e.second.second);
    ans &= (e.first == mini);
  }
  if (ans) {
    for (int i = 0; i < (int)input.size(); i++) {
      cout << max(Gm[input[i].first][input[i].second], 1) << " ";
    }
    cout << endl;
  } else
    cout << -1 << endl;
  return 0;
}