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 N = (int)5e3 + 10;
int n, m, edge[N];
vector<pair<int, int> > adj[N];
pair<int, pair<int, int> > query[N];
vector<int> path[N];
bool visited[N];
bool find_path(int u, int v, int id) {
  if (u == v) return true;
  visited[u] = true;
  for (auto it : adj[u])
    if (!visited[it.second]) {
      path[id].push_back(it.first);
      if (find_path(it.second, v, id)) return true;
      path[id].pop_back();
    }
  visited[u] = false;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(pair<int, int>(i, v));
    adj[v].push_back(pair<int, int>(i, u));
  }
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    query[i] = pair<int, pair<int, int> >(w, pair<int, int>(u, v));
    memset(visited, false, sizeof(visited));
    find_path(u, v, i);
    for (auto it : path[i]) {
      edge[it] = max(edge[it], w);
    }
  }
  for (int i = 1; i <= m; i++) {
    int temp = INT_MAX;
    for (auto it : path[i]) temp = min(temp, edge[it]);
    if (temp != query[i].first) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i < n; i++) {
    if (edge[i] == 0) edge[i] = 911218;
    cout << edge[i] << ' ';
  }
  return 0;
}