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;
struct arco {
  int x, y, in;
  bool operator<(const arco& a) const& { return in < a.in; }
};
int n, m;
arco archi[10000];
vector<arco> adj[10000], sor;
bool dfs(int a, int p, int arr, int mim) {
  if (a == arr) return 1;
  for (auto i : adj[a]) {
    if (i.y != p) {
      if (dfs(i.y, a, arr, mim)) {
        archi[i.in].in = mim;
        return 1;
      }
    }
  }
  return 0;
}
int cec(int a, int p, int arr) {
  if (a == arr) return 1000000000;
  for (auto i : adj[a]) {
    if (i.y != p) {
      int x = cec(i.y, a, arr);
      if (x != 0) {
        return min(archi[i.in].in, x);
      }
    }
  }
  return 0;
}
int main() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    archi[i].in = 1;
    int x, y;
    cin >> x >> y;
    adj[x].push_back({x, y, i});
    adj[y].push_back({y, x, i});
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int x, y, p;
    cin >> x >> y >> p;
    sor.push_back({x, y, p});
  }
  sort(sor.begin(), sor.end());
  for (int i = 0; i < m; i++) {
    dfs(sor[i].x, -1, sor[i].y, sor[i].in);
  }
  for (int i = 0; i < m; i++) {
    if (cec(sor[i].x, -1, sor[i].y) != sor[i].in) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    cout << archi[i].in << " ";
  }
}