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 <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
using ll = long long int;
using ld = long double;
using vi = vector<int>;
using mii = map<int, int>;
using pii = pair<int, int>;
const int inf = 1e9;
const int N = 100005;
const int MAXN = 5005;
const ll mod = (ll)1e9 + 7;
vector<pair<int, int>> g[MAXN];
pair<int, int> p[MAXN];
int tin[MAXN];
int tout[MAXN];
int ans[MAXN];
int tim;
bool used[MAXN];
void dfs(int v) {
  used[v] = 1;
  tin[v] = tim++;
  for (auto to : g[v]) {
    if (used[to.first]) continue;
    p[to.first] = {v, to.second};
    dfs(to.first);
  }
  tout[v] = tim++;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int aa, bb;
    cin >> aa >> bb;
    aa--;
    bb--;
    g[aa].push_back({bb, i});
    g[bb].push_back({aa, i});
  }
  for (int i = 0; i < n - 1; i++) {
    ans[i] = 1e6;
  }
  int m;
  cin >> m;
  vector<pair<int, pair<int, int>>> q;
  for (int i = 0; i < m; i++) {
    int aa, bb, cc;
    cin >> aa >> bb >> cc;
    aa--;
    bb--;
    q.push_back({cc, {aa, bb}});
  }
  sort(q.begin(), q.end());
  dfs(0);
  for (auto qu : q) {
    int x, y;
    tie(x, y) = qu.second;
    while (!(tin[x] <= tin[y] && tout[x] >= tin[y])) {
      ans[p[x].second] = qu.first;
      x = p[x].first;
    }
    while (!(tin[y] <= tin[x] && tout[y] >= tin[x])) {
      ans[p[y].second] = qu.first;
      y = p[y].first;
    }
  }
  for (auto qu : q) {
    int x, y;
    tie(x, y) = qu.second;
    int mn = 1e6;
    while (!(tin[x] <= tin[y] && tout[x] >= tin[y])) {
      mn = min(mn, ans[p[x].second]);
      x = p[x].first;
    }
    while (!(tin[y] <= tin[x] && tout[y] >= tin[x])) {
      mn = min(mn, ans[p[y].second]);
      y = p[y].first;
    }
    if (mn != qu.first) {
      cout << -1 << '\n';
      return 0;
    }
  }
  for (int i = 0; i < n - 1; i++) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
  return 0;
}