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 cmp(pair<pair<int, int>, int> a, pair<pair<int, int>, int> b) {
  return a.second > b.second;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<pair<int, int> > vvv;
  vector<int> adj[n + 5];
  int f[n + 5][n + 5];
  memset(f, 0, sizeof(f));
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    vvv.push_back({x, y});
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  int m;
  cin >> m;
  pair<pair<int, int>, int> p[m + 5];
  for (int i = 1; i <= m; i++)
    cin >> p[i].first.first >> p[i].first.second >> p[i].second;
  sort(p + 1, p + m + 1, cmp);
  int vis[n + 5];
  int par[n + 5];
  for (int i = 1; i <= m; i++) {
    memset(vis, -1, sizeof(vis));
    memset(par, -1, sizeof(par));
    queue<int> q;
    int start = p[i].first.first, end = p[i].first.second;
    q.push(start);
    vis[start] = 0;
    while (!q.empty()) {
      int cur = q.front();
      q.pop();
      for (int i = 0; i < adj[cur].size(); i++) {
        int next = adj[cur][i];
        if (vis[next] == -1) {
          par[next] = cur;
          vis[next] = vis[cur] + 1;
          q.push(next);
        }
      }
      if (vis[end] != -1) break;
    }
    int cur = end;
    bool y = 0;
    while (cur != start) {
      if (f[cur][par[cur]] <= p[i].second) y = 1;
      if (f[par[cur]][cur] <= p[i].second) y = 1;
      f[cur][par[cur]] = max(f[cur][par[cur]], p[i].second);
      f[par[cur]][cur] = max(f[par[cur]][cur], p[i].second);
      cur = par[cur];
    }
    if (!y) {
      cout << -1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < vvv.size(); i++) {
    if (f[vvv[i].first][vvv[i].second] == 0)
      cout << 1 << " ";
    else
      cout << f[vvv[i].first][vvv[i].second] << " ";
  }
  return 0;
}