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, ans[5005];
pair<int, int> pre[5005];
vector<pair<int, int>> adj[5005], q[5005];
void dfs(int u, int par) {
  for (pair<int, int> nxt : adj[u]) {
    int v = nxt.first;
    if (v == par) continue;
    pre[v] = {u, nxt.second};
    dfs(v, u);
  }
}
void update(int u, int v, int val) {
  while (v != u) {
    ans[pre[v].second] = max(ans[pre[v].second], val);
    v = pre[v].first;
  }
}
int check(int u, int v, int val) {
  int cur = INT_MAX;
  while (v != u) {
    cur = min(cur, ans[pre[v].second]);
    v = pre[v].first;
  }
  return cur == val;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back({v, i});
    adj[v].push_back({u, i});
    ans[i] = 1;
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int u, v, val;
    cin >> u >> v >> val;
    q[u].push_back({v, val});
  }
  for (int i = 1; i <= n; i++) {
    if (q[i].empty()) continue;
    pre[i] = {i, 0};
    dfs(i, i);
    for (pair<int, int> cur : q[i]) {
      update(i, cur.first, cur.second);
    }
  }
  bool ok = true;
  for (int i = 1; i <= n; i++) {
    if (q[i].empty()) continue;
    pre[i] = {i, 0};
    dfs(i, i);
    for (pair<int, int> cur : q[i]) {
      ok &= check(i, cur.first, cur.second);
    }
  }
  if (ok) {
    for (int i = 0; i < n - 1; i++) {
      cout << ans[i] << (i + 1 < n - 1 ? ' ' : '\n');
    }
  } else {
    cout << -1 << '\n';
  }
  return 0;
}