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, m;
vector<vector<int>> adjlist;
vector<int> trace;
const int nax = 5005;
const int INF = 1e9;
int ans[nax];
struct P {
  int a, b, f;
  bool operator<(const P& other) const { return f < other.f; }
};
P passenger[nax];
int edge[nax][nax];
void dfs(int v, int p = -1) {
  trace[v] = p;
  for (int& c : adjlist[v]) {
    if (c == p) continue;
    dfs(c, v);
  }
}
void Destroy(vector<int>& v) {
  v.clear();
  v.shrink_to_fit();
}
void bye() {
  cout << -1;
  exit(0);
}
void solve() {
  cin >> n;
  adjlist.resize(n + 1);
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    adjlist[u].push_back(v);
    adjlist[v].push_back(u);
    edge[u][v] = i;
    edge[v][u] = i;
  }
  cin >> m;
  for (int i = 1; i < n; ++i) ans[i] = 1;
  for (int i = 1; i <= m; ++i) {
    cin >> passenger[i].a >> passenger[i].b >> passenger[i].f;
  }
  sort(passenger + 1, passenger + m + 1);
  for (int i = 1; i <= m; ++i) {
    int start = passenger[i].a, finish = passenger[i].b;
    int mini = passenger[i].f;
    Destroy(trace);
    trace.resize(nax);
    dfs(start);
    vector<int> path;
    for (int j = finish; j != start; j = trace[j]) {
      path.push_back(j);
    }
    path.push_back(start);
    for (int id = 0; id < ((int)path.size()) - 1; ++id) {
      int get_id = edge[path[id]][path[id + 1]];
      ans[get_id] = max(ans[get_id], mini);
    }
  }
  for (int i = 1; i <= m; ++i) {
    int start = passenger[i].a, finish = passenger[i].b;
    int mini = passenger[i].f;
    Destroy(trace);
    trace.resize(nax);
    dfs(start);
    vector<int> path;
    for (int j = finish; j != start; j = trace[j]) {
      path.push_back(j);
    }
    path.push_back(start);
    int compare = INF;
    for (int id = 0; id < ((int)path.size()) - 1; ++id) {
      int get_id = edge[path[id]][path[id + 1]];
      compare = min(compare, ans[get_id]);
    }
    if (compare != mini) {
      bye();
    }
  }
  for (int i = 1; i < n; ++i) cout << ans[i] << ' ';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}