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 dfs(int curr, const vector<vector<pair<int, int>>>& gr,
         vector<pair<int, int>>& parent, int par = -1, int edge = -1) {
  parent[curr] = {par, edge};
  for (const auto& next : gr[curr]) {
    if (next.first != par) {
      dfs(next.first, gr, parent, curr, next.second);
    }
  }
}
int main() {
  int n;
  cin >> n;
  vector<vector<pair<int, int>>> gr(n);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    gr[u].push_back({v, i});
    gr[v].push_back({u, i});
  }
  vector<int> f(n - 1, 1);
  vector<vector<pair<int, int>>> parent(n, vector<pair<int, int>>(n, {-1, -1}));
  for (int i = 0; i < n; ++i) {
    dfs(i, gr, parent[i]);
  }
  int m;
  cin >> m;
  vector<array<int, 3>> queries(m);
  for (auto& q : queries) {
    cin >> q[1] >> q[2] >> q[0];
    --q[1];
    --q[2];
  }
  sort(queries.begin(), queries.end());
  for (const auto& q : queries) {
    int curr = q[1];
    while (parent[q[2]][curr].first != -1) {
      f[parent[q[2]][curr].second] = q[0];
      curr = parent[q[2]][curr].first;
    }
  }
  for (const auto& q : queries) {
    int res = 1000 * 1000 * 1000;
    int curr = q[1];
    while (parent[q[2]][curr].first != -1) {
      res = min(res, f[parent[q[2]][curr].second]);
      curr = parent[q[2]][curr].first;
    }
    if (res != q[0]) {
      cout << "-1\n";
      return 0;
    }
  }
  for (int i = 0; i < n - 1; ++i) {
    cout << f[i] << " ";
  }
  cout << "\n";
}