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;
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << '\n';
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
const int N = 5e3 + 3;
int n, m, d[N][N], p[N];
bool vis[N];
vector<int> g[N];
vector<pair<int, int> > edges;
vector<pair<int, pair<int, int> > > q;
bool valid;
void fil(int st, int en, int col) {
  memset(p, -1, sizeof p);
  queue<int> q;
  q.push(st);
  p[st] = 0;
  while (!q.empty()) {
    int v = q.front();
    q.pop();
    if (v == en) break;
    for (auto nb : g[v])
      if (p[nb] == -1) {
        p[nb] = v;
        q.push(nb);
      }
  }
  (st, en);
  bool done = false;
  for (int cur = en; cur != st; cur = p[cur]) {
    if (d[cur][p[cur]] == 0) {
      (cur, p[cur], col);
      d[cur][p[cur]] = d[p[cur]][cur] = col, done = true;
    } else {
      assert(d[cur][p[cur]] >= col);
      if (d[cur][p[cur]] == col) done = true;
    }
  }
  if (!done) valid = false;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
    edges.push_back(make_pair(a, b));
  }
  cin >> m;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    q.push_back(make_pair(c, make_pair(a, b)));
  }
  sort((q).begin(), (q).end());
  reverse((q).begin(), (q).end());
  valid = true;
  for (auto e : q) {
    int st = e.second.first, en = e.second.second, col = e.first;
    (e);
    if (st == en) {
      cout << -1 << '\n';
      return 0;
    }
    fil(st, en, col);
    if (!valid) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (auto e : edges)
    cout << (d[e.first][e.second] != 0 ? d[e.first][e.second] : 1000000) << " ";
  return 0;
}