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 MAX = 1000000;
int main() {
  int n, m;
  cin >> n;
  vector<vector<int>> adj(n, vector<int>());
  vector<pair<int, int>> edges;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
    edges.push_back(pair<int, int>(u, v));
  }
  vector<vector<int>> parent(n, vector<int>(n, -1));
  for (int i = 0; i < n; ++i) {
    queue<int> q;
    q.push(i);
    parent[i][i] = i;
    while (!q.empty()) {
      int c = q.front();
      q.pop();
      for (int child : adj[c]) {
        if (parent[i][child] == -1) {
          parent[i][child] = c;
          q.push(child);
        }
      }
    }
  }
  vector<vector<int>> cost(n, vector<int>(n, -1));
  cin >> m;
  vector<int> a(m), b(m), g(m);
  for (int i = 0; i < m; ++i) {
    int bi, ai, gi;
    cin >> ai >> bi >> gi;
    --ai, --bi;
    a[i] = ai;
    b[i] = bi;
    g[i] = gi;
    while (bi != ai) {
      int p = parent[ai][bi];
      cost[p][bi] = cost[bi][p] = max(cost[p][bi], g[i]);
      bi = p;
    }
  }
  for (int i = 0; i < m; ++i) {
    int min_val = MAX;
    int ai = a[i];
    int bi = b[i];
    while (bi != ai) {
      int p = parent[ai][bi];
      min_val = min(min_val, cost[p][bi]);
      bi = p;
    }
    if (min_val != g[i]) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (auto iter : edges) {
    int u = iter.first;
    int v = iter.second;
    if (cost[u][v] == -1)
      cout << MAX << " ";
    else
      cout << cost[u][v] << " ";
  }
  cout << endl;
  return 0;
}