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;
bool dfs(int cur, int par, int target, const vector<vector<pair<int, int>>>& g,
         vector<int>& ans, int val, bool& ok) {
  if (cur == target) return true;
  for (auto child : g[cur])
    if (child.first != par) {
      if (dfs(child.first, cur, target, g, ans, val, ok)) {
        if (ans[child.second] == -1) ans[child.second] = val;
        if (ans[child.second] == val) ok = true;
        return true;
      }
    }
  return false;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<pair<int, int>>> g(n);
  for (int i = 0; i + 1 < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].emplace_back(b, i);
    g[b].emplace_back(a, i);
  }
  vector<int> ans(n - 1, -1);
  vector<vector<int>> query;
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    query.push_back({c, a, b});
  }
  sort(query.rbegin(), query.rend());
  for (int i = 0; i < m; i++) {
    bool ok = false;
    dfs(query[i][1], -1, query[i][2], g, ans, query[i][0], ok);
    if (!ok) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i + 1 < n; i++) {
    if (i > 0) cout << " ";
    cout << (ans[i] == -1 ? (int)1e6 : ans[i]);
  }
  cout << endl;
  return 0;
}