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 T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class T>
ostream &operator<<(ostream &os, set<T> S) {
  os << "{ ";
  for (auto s : S) os << s << " ";
  return os << "}";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class L, class R>
ostream &operator<<(ostream &os, map<L, R> M) {
  os << "{ ";
  for (auto m : M) os << "(" << m.first << ":" << m.second << ") ";
  return os << "}";
}
const int N = 5010;
vector<pair<int, int> > graf[N];
int h[N], st[N], parent[N];
int A[N], B[N], C[N];
void dfs(int node, int par) {
  h[node] = h[par] + 1;
  parent[node] = par;
  for (auto v : graf[node]) {
    if (v.first != par) {
      st[v.first] = v.second;
      dfs(v.first, node);
    }
  }
}
int v[N];
void init(int a, int b, int c) {
  if (h[a] < h[b]) swap(a, b);
  while (h[a] != h[b]) {
    v[st[a]] = max(v[st[a]], c);
    a = parent[a];
  }
  while (a != b) {
    v[st[a]] = max(v[st[a]], c);
    v[st[b]] = max(v[st[b]], c);
    a = parent[a];
    b = parent[b];
  }
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int n;
  cin >> n;
  for (int i = 1; i < (n); ++i) {
    v[i] = 1;
    int x, y;
    cin >> x >> y;
    graf[x].push_back({y, i});
    graf[y].push_back({x, i});
  }
  dfs(1, 0);
  function<int(int, int)> f = [&](int u, int va) {
    int inf = 1e9;
    int ans = inf + 10;
    if (h[u] < h[va]) swap(u, va);
    while (h[u] != h[va]) {
      ans = min(ans, v[st[u]]);
      u = parent[u];
    }
    while (u != va) {
      ans = min(ans, v[st[u]]);
      ans = min(ans, v[st[va]]);
      u = parent[u];
      va = parent[va];
    }
    return ans;
  };
  int m;
  cin >> m;
  for (int i = 1; i < (m + 1); ++i) {
    cin >> A[i] >> B[i] >> C[i];
    init(A[i], B[i], C[i]);
  }
  for (int i = 1; i < (m + 1); ++i) {
    if (f(A[i], B[i]) != C[i]) {
      cout << -1;
      return 0;
    }
  }
  for (int i = 1; i < (n); ++i) cout << v[i] << " ";
  return 0;
}