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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;
mt19937_64 mt_rand(chrono::system_clock::now().time_since_epoch().count());
template <typename T1, typename T2>
inline bool upmax(T1& a, T2 b) {
  return (a < b ? (a = b, true) : false);
}
template <typename T1, typename T2>
inline bool upmin(T1& a, T2 b) {
  return (b < a ? (a = b, true) : false);
}
const int maxn = (int)5e5 + 100;
const int maxlog = 21;
const int base = 998244353;
const double eps = (double)1e-9;
const double PI = acos(-1.);
const int inf = 2e9;
const long long llinf = 1e18;
vector<pair<int, int>> g[maxn];
int res[maxn];
bool dfs(int v, int p, int e, int c) {
  if (v == e) {
    return true;
  }
  for (auto [to, id] : g[v]) {
    if (to == p) continue;
    if (dfs(to, v, e, c)) {
      upmax(res[id], c);
      return true;
    }
  }
  return false;
}
void upmax(int a, int b, int c) { dfs(a, a, b, c); }
bool dfs2(int v, int p, int e, int& cur) {
  if (v == e) return true;
  for (auto [to, id] : g[v]) {
    if (to == p) continue;
    if (dfs2(to, v, e, cur)) {
      upmin(cur, res[id]);
      return true;
    }
  }
  return false;
}
bool check(int a, int b, int c) {
  int cur = inf;
  dfs2(a, a, b, cur);
  return cur == c;
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    res[i] = 1;
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back({b, i});
    g[b].push_back({a, i});
  }
  int m;
  cin >> m;
  vector<tuple<int, int, int>> q;
  for (int i = 0; i < m; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    q.emplace_back(c, a, b);
  }
  sort((q).begin(), (q).end());
  for (auto [c, a, b] : q) {
    upmax(a, b, c);
  }
  for (auto [c, a, b] : q) {
    if (!check(a, b, c)) {
      cout << -1;
      return;
    }
  }
  for (int i = 1; i < n; i++) {
    cout << res[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}