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;
int n;
int y;
vector<vector<int>> adj;
vector<pair<int, int>> vec;
vector<pair<int, pair<int, int>>> q;
int val[5005][5005];
bool boolum;
bool dfs(int u, int v, int p, int w) {
  if (u == v) {
    return true;
  }
  for (auto node : adj[u]) {
    if (node != p) {
      if (dfs(node, v, u, w)) {
        if (val[node][u] == 0 || val[node][u] == w) {
          boolum = true;
        }
        if (val[node][u] == 0) {
          val[node][u] = w;
          val[u][node] = w;
        }
        return true;
      }
    }
  }
  return false;
}
int main() {
  cin >> n;
  adj.resize(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
    vec.push_back({u, v});
  }
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    q.push_back({w, {u, v}});
  }
  sort(q.begin(), q.end());
  reverse(q.begin(), q.end());
  for (int i = 0; i < m; i++) {
    int u = q[i].second.first;
    int v = q[i].second.second;
    int w = q[i].first;
    bool booly;
    boolum = false;
    booly = dfs(u, v, u, w);
    if (boolum == false) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    int x = val[vec[i].first][vec[i].second];
    if (x == 0) {
      x = pow(10, 6);
    }
    cout << x << " ";
  }
  return 0;
}